Calculus/Derivatives

Notation
We denote the derivative of a function $$f$$ at a number $$a$$ as $$f'(a)\,\!$$.

Definition
The derivative of a function $$f$$ at a number $$a$$ a is given by the following limit (if it exists): $$f'(a)=\lim_{h\rightarrow 0}\frac{f(a+h)-f(a)}{h}$$ An analagous equation can be defined by letting $$x=(a+h)$$. Then $$h=(x-a)$$, which shows that when $$x$$ approaches $$a$$, $$h$$ approaches $$0$$: $$f'(a)=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}$$

As the slope of a tangent line
Given a function $$y=f(x)\,\!$$, the derivative $$f '(a)\,\!$$ can be understood as the slope of the tangent line to $$f(x)$$ at $$x=a$$:

Example
Find the equation of the tangent line to $$y=x^2$$ at $$x=1$$.

Solution
To find the slope of the tangent, we let $$y=f(x)$$ and use our first definition: $$f'(a)=\lim_{h\rightarrow 0}\frac{f(a+h)-f(a)}{h}\Rightarrow \lim_{h\rightarrow 0}\frac{h}\Rightarrow \lim_{h\rightarrow 0}\frac{h}\Rightarrow \lim_{h\rightarrow 0}\frac{h}\Rightarrow \lim_{h\rightarrow 0}{\color{Blue}(2a+h)}$$ It can be seen that as $$h$$ approaches $$0$$, we are left with $$f'(a)={\color{Blue}2a}\,\!$$. If we plug in $$1$$ for $$a$$: $$f'({\color{Red}1})=2({\color{Red}1})\Rightarrow {\color{Red}2}$$ So our preliminary equation for the tangent line is $$y={\color{Red}2}x+b$$. By plugging in our tangent point $$(1,1)$$ to find $$b$$, we can arrive at our final equation: $${\color{Red}1}=2({\color{Red}1})+b\Rightarrow -1=b$$ So our final equation is $$y=2x-1\,\!$$.

As a rate of change
The derivative of a function $$f(x)$$ at a number $$a$$ can be understood as the instantaneous rate of change of $$f(x)$$ when $$x=a$$.

The derivative as a function
So far we have only examined the derivative of a function $$f$$ at a certain number $$a$$. If we move from the constant $$a$$ to the variable $$x$$, we can calculate the derivative of the function as a whole, and come up with an equation that represents the derivative of the function $$f$$ at any arbitrary $$x$$ value. For clarification, the derivative of $$f$$ at $$a$$ is a number, whereas the derivative of $$f$$ is a function.

Notation
Likewise to the derivative of $$f$$ at $$a$$, the derivative of the function $$f(x)$$ is denoted $$f'(x)\,\!$$.

Definition
The derivative of the function $$f$$ is defined by the following limit: $$f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$$ Also, $$f'(x)=\lim_{h\rightarrow x}\frac{f(x)-f(h)}{x-h}$$ or $$f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x-h)}{2h}$$

d(x^n)/dx
Consider the sequences:

$$y = x^2;\ y + \Delta y = (x + h)^2 = x^2 + 2xh + h^2$$

$$y = x^3;\ y + \Delta y = (x + h)^3 = x^3 + 3x^2h + 3xh^2 + h^3$$

$$y = x^4;\ y + \Delta y = (x + h)^4 = x^4 + 4x^3h + 6x^2h^2 + 4xh^3 + h^4$$

$$y = x^5;\ y + \Delta y = (x + h)^5 = x^5 + 5x^4h + 10x^3h^2 + 10x^2h^3 + 5xh^4 + h^5$$

$$y = x^6;\ y + \Delta y = (x + h)^6 = x^6 + 6x^5h\ +$$ many terms containing $$h^2$$

$$y = x^n;\ y + \Delta y = (x + \Delta x)^n = x^n + nx^{n-1}\Delta x\ +$$ many terms containing $$(\Delta x)^2$$

$$\Delta y = nx^{n-1}\Delta x\ +$$ many terms containing $$(\Delta x)^2$$

Therefore:

$$\frac{\Delta y}{\Delta x} = nx^{n-1}\ +$$ many terms with each and every term containing $$\Delta x$$

$$\frac{dy}{dx} = \lim_{\Delta x\rightarrow 0}\frac{\Delta y}{\Delta x}$$

$$\frac{d(x^n)}{dx} = nx^{n-1}$$ read as "derivative with respect to $$x$$ of $$x$$ to the power $$n$$."

Later it will be shown that this is valid for all real $$n$$, positive or negative, integer or fraction.

Examples
$$\frac{d(x^4)}{dx} = 4x^{(4-1)} = 4x^3$$

$$\frac{d(7x^5)}{dx} = 7\frac{d(x^5)}{dx} = 7(5x^4) = 35x^4$$

$$\frac{d(p^3)}{dx} = \frac{d(p^3)}{dp}$$$⋅$$$\frac{dp}{dx} = 3p^2$$$⋅$$$\frac{dp}{dx}$$

$$\frac{d(2t^2 - 7T^3 + x^7)}{dx} = 4t$$$⋅$$$\frac{dt}{dx} - 7(3T^2)$$$⋅$$$\frac{dT}{dx} + 7x^6$$

$$\frac{d(x^{\frac{1}{3}})}{dx} = \frac{1}{3}x^{\frac{1}{3}-1} = \frac{1}{3}x^{-\frac{2}{3}}$$

Derivative of cubic function
In the diagram there is a stationary point at $$x = p.$$ When $$x = p,$$ there are exactly 2 values of $$x$$ that produce $$f(x) = f(p).$$

Aim of this section is to derive the condition that produces exactly 2 values of $$x.$$

See Cubic function as product of linear function and quadratic.

The associated quadratic when $$x = p:$$

$$A = a$$

$$B = Ap + b = ap+b$$

$$C = Bp + c = app + bp + c.$$

Divide $$g(x) = ax^2 + (ap+b)x + (ap^2 + bp + c)$$ by $$(x-p).$$

This division gives a quotient of $$ax + (2ap+b)$$ and remainder of $$h(p) = 3ap^2 + 2bp + c.$$

Factor $$(x-p)$$ divides $$g(x)$$ exactly. Therefore, remainder $$h(p) = 0,$$ the desired condition.

When $$x = p,\ h(x) = 3ax^2 + 2bx + c,$$ the derivative of $$f(x).$$

Derivative of quartic function
The quartic function: $$y = f(x) = ax^4 + bx^3 + cx^2 + dx + e\ \dots\ (1)$$

In $$(1)$$ substitute $$(p + q)$$ for $$x\ \dots\ (2a).$$

In $$(1)$$ substitute $$(p - q)$$ for $$x\ \dots\ (3a).$$

$$(2a) + (3a):\ + 2apppp + 12appqq + 2aqqqq + 2bppp + 6bpqq + 2cpp + 2cqq + 2dp + 2e\ \dots\ (4)$$

$$(2a) - (3a):\ + 8apppq + 8apqqq + 6bppq + 2bqqq + 4cpq + 2dq\ \dots\ (5)$$

Reduce (4) and (5) and substitute Q for qq:

$$+ apppp + 6appQ + aQQ + bppp + 3bpQ + cpp + cQ + dp + e\ \dots\ (4a)$$

$$+ 4appp + 4apQ + 3bpp + bQ + 2cp + d\ \dots\ (5a)$$

Combine (4a) and (5a) to eliminate p and produce a function in Q:

$$C_{11}Q^{11} + C_{10}Q^{10} + C_9Q^9 +\ \cdots\ + C_2Q^2 + C_1Q + C_0$$ where:

$$C_{11} = (- 65536aaaaaaaaaabb)$$

$$C_{10} = (- 131072aaaaaaaaaabd - 229376aaaaaaaaabbc + 94208aaaaaaaabbbb)$$

$$\cdots\cdots$$

$$C_2 = (- 8192aaaaaabdeeee + 20480aaaaaacddeee \cdots - 288aabbbccccddd + 32aabbccccccdd)$$

$$C_1 = (- 4096aaaaaaddeeee - 4096aaaaabcdeeee \cdots + 64aabbcccccdde - 20aabbccccdddd)$$

Coefficient of interest is $$C_0$$ which is in fact the value $$Rs - Sr.$$ See Equal Roots of Quartic Function.

If $$C_0 == 0,\ q = Q = 0$$ is a solution and $$f(x)$$ contains at least 2 roots of format $$p+0, p-0.$$ In other words 2 equal roots where $$x = p.$$

If $$Q == 0,\ (4a)$$ and $$(5a)$$ become:

$$ap^4 + bp^3 + cp^2 + dp + e\ \dots\ (4b)$$

$$4ap^3 + 3bp^2 + 2cp + d\ \dots\ (5b)$$

$$(4b)$$ is equivalent to $$f(x)$$ and $$(5b)$$ is derivative of $$(4b).$$

Product Rule
Let $$y = u$$$⋅$$$v$$ where $$u = U(x);\ v = V(x)$$

$$y + \Delta y = (u + \Delta u)$$$⋅$$$(v + \Delta v) = u$$$⋅$$$v + u$$$⋅$$$\Delta v + v$$$⋅$$$\Delta u + \Delta u$$$⋅$$$\Delta v$$

$$\Delta y = u$$$⋅$$$\Delta v + v$$$⋅$$$\Delta u + \Delta u$$$⋅$$$\Delta v$$

$$\frac{\Delta y}{\Delta x} = u$$$⋅$$$\frac{\Delta v}{\Delta x} + v$$$⋅$$$\frac{\Delta u}{\Delta x} + \frac{(\Delta u)(\Delta v)}{\Delta x}$$

$$\frac{dy}{dx} = u$$$⋅$$$\frac{dv}{dx} + v$$$⋅$$$\frac{du}{dx}$$

Examples
Let $$y = x^{\frac{m}{n}}.$$ Calculate $$\frac{dy}{dx}$$

$$y^n = x^m$$

Differentiate both sides.

$$ny^{n-1}\cdot\frac{dy}{dx} = mx^{m-1}$$

$$\frac{dy}{dx} = \frac{mx^{m-1}}{ ny^{n-1}    }$$ $$ = \frac{mx^{m-1}}{ n(x^{\frac{m}{n}})^{n-1}    }$$ $$ = \frac{mx^{m-1}}{ nx^{m - \frac{m}{n}}    }$$ $$= \frac{m}{n}\cdot x^{m - 1 - m + \frac{m}{n} }$$ $$= \frac{m}{n}\cdot x^{\frac{m}{n} - 1}$$

This shows that $$\frac{d(x^n)}{dx}$$ as above is valid when $$n$$ is a positive fraction.

Let $$y = x^{-n} = \frac{1}{x^n}$$. Calculate $$\frac{dy}{dx}$$.

$$yx^n = 1$$

Differentiate both sides.

$$ynx^{n-1} + x^n\frac{dy}{dx} = 0$$

$$\frac{dy}{dx} = -\frac{ynx^{n-1}}{x^n} = -\frac{nx^{n-1}}{x^{n}x^n} = -nx^{n-1}x^{-2n} = -nx^{-n-1}$$

This shows that $$\frac{d(x^n)}{dx}$$ as above is valid for negative $$n$$.

Let $$y = x^{-\frac{m}{n}}$$. Calculate $$\frac{dy}{dx}$$.

$$y^n = x^{-m}$$

Differentiate both sides.

$$ny^{n-1}\frac{dy}{dx} = -mx^{-m-1}$$

$$\frac{dy}{dx} = -\frac{mx^{-m-1}}{ny^{n-1}} = -\frac{mx^{-m-1}}{n(x^{-\frac{m}{n}})^{n-1}}$$ $$ = -\frac{mx^{-m-1}}{nx^{-m+\frac{m}{n}}}$$ $$ = -\frac{m}{n}x^{-m-1+m-\frac{m}{n}}$$ $$ = -\frac{m}{n}x^{-\frac{m}{n}-1}$$

This shows that $$\frac{d(x^n)}{dx}$$ as above is valid when $$n$$ is a negative fraction.

Quotient rule
Used where $$y = \frac{U(x)}{V(x)}$$

$$y = \frac{u}{v}$$

$$yv = u$$

$$yv' + vy' = u'$$

$$vy' = u' - yv' = u' - \frac{uv'}{v}$$

$$y' = \frac{u'}{v} - \frac{uv'}{v^2} = \frac{vu'}{v^2} - \frac{uv'}{v^2} = \frac{vu' - uv'}{v^2}$$

sine(x)
$$y = \sin (x)$$

$$y + \Delta y = \sin (x + \Delta x) = \sin(x)\cos(\Delta x) + \cos(x)\sin(\Delta x)$$

$$\Delta y = \sin(x)\cos(\Delta x) + \cos(x)\sin(\Delta x) - \sin(x)$$

$$\frac{\Delta y}{\Delta x} = \sin(x)$$$⋅$$$\frac{\cos(\Delta x) - 1}{\Delta x} + \cos(x)$$$⋅$$$\frac{\sin(\Delta x)}{\Delta x} $$

The value $$\frac{\cos(\Delta x) - 1}{\Delta x}$$:

$$\lim_{\Delta x\rightarrow 0}\frac{\cos(\Delta x)-1}{\Delta x} = \lim_{\Delta x\rightarrow 0}\frac{-\sin(\Delta x)}{1} = 0$$ by L'Hôpital's rule.

The value $$\frac{\sin(\Delta x)}{\Delta x}$$:

$$\lim_{\Delta x\rightarrow 0}\frac{\sin(\Delta x)}{\Delta x} = \lim_{\Delta x\rightarrow 0}\frac{\cos(\Delta x)}{1} = 1$$ by L'Hôpital's rule.

$$\frac{dy}{dx} = \sin(x)$$$⋅$$$0 + \cos(x)$$$⋅$$$1 = \cos(x)$$

Proof of 2 limits
$$\lim_{\theta\rightarrow 0}\frac{\sin(\theta) }{\theta }:$$

In the diagram $$OA = OB = 1.$$ $$OC = OD = \cos(\theta).$$ $$CB = \sin(\theta).$$

Let $$S$$ be the area of a sector of a circle. Then $$\frac{S}{\pi r^2} = \frac{\theta}{2\pi}$$ and $$S = \frac{\theta r^2}{2}.$$

Area of sector $$COD = \frac{\theta \cos^2(\theta)}{2}.$$

Area of triangle $$OCB = \frac{\sin(\theta) \cos(\theta)}{2}.$$

Area of sector $$AOB = \frac{\theta \cdot 1^2}{2} = \frac{\theta}{2}.$$

Therefore $$\frac{\theta}{2}\ \ge\ \frac{\sin(\theta) \cos(\theta)}{2}\ \ge\ \frac{\theta \cos^2(\theta)}{2}.$$

$$\theta\ \ge\ \sin(\theta) \cos(\theta)\ \ge\ \theta \cos^2(\theta).$$

$$\frac{\theta}{\theta \cos(\theta)}\ \ge\ \frac{\sin(\theta) \cos(\theta)}{\theta \cos(\theta)}\ \ge\ \frac{\theta \cos^2(\theta)}{\theta \cos(\theta)}$$

$$\frac{1}{ \cos(\theta)}\ \ge\ \frac{\sin(\theta) }{\theta }\ \ge\ \cos(\theta)$$

$$\lim_{\theta\rightarrow 0}  \frac{1}{ \cos(\theta)}\ \ge\ \lim_{\theta\rightarrow 0}\frac{\sin(\theta) }{\theta }\ \ge\ \lim_{\theta\rightarrow 0}\cos(\theta)  $$

$$1 \ \ge\ \lim_{\theta\rightarrow 0}\frac{\sin(\theta) }{\theta }\ \ge\ 1 $$

Therefore $$\lim_{\theta\rightarrow 0}\frac{\sin(\theta) }{\theta } = 1.$$

$$\lim_{\theta\rightarrow 0} \frac{\cos(\theta) - 1}{\theta}:$$

$$\frac{\cos(\theta) - 1}{\theta}$$ $$= (-1)\cdot \frac{1 - \cos(\theta)}{\theta}$$ $$= (-1)\cdot \frac{1 - \cos(\theta)}{\theta} \cdot \frac{1 + \cos(\theta)}{1 + \cos(\theta)} $$ $$= (-1)\cdot \frac{ \sin(\theta)}{\theta} \cdot \frac{ \sin(\theta)}{1 + \cos(\theta)} $$

$$\lim_{\theta\rightarrow 0} \frac{\cos(\theta) - 1}{\theta}$$ $$=\lim_{\theta\rightarrow 0} (-1)\cdot \frac{ \sin(\theta)}{\theta} \cdot \frac{ \sin(\theta)}{1 + \cos(\theta)} $$ $$=(-1)\cdot \lim_{\theta\rightarrow 0} \frac{ \sin(\theta)}{\theta} \cdot \lim_{\theta\rightarrow 0} \frac{ \sin(\theta)}{1 + \cos(\theta)} $$ $$=(-1)\cdot 1 \cdot \frac{ 0}{1 + 1}  = 0.$$

cosine(x)
$$y = \cos(x) = \sqrt{1 - \sin^2(x)}$$

$$y^2 = 1 - \sin(x)$$$⋅$$$\sin(x)$$

Differentiate both sides:

$$2y$$$⋅$$$\frac{dy}{dx} = -(2\sin(x)\cos(x))$$

$$\frac{dy}{dx} = -\frac{\sin(x)\cos(x)}{\cos(x)} = -\sin(x).$$

tan(x)
$$y = \tan(x) = \frac{\sin(x)}{\cos(x)}$$

$$y\cos(x) = \sin(x)$$

Differentiate both sides:

$$y(-\sin(x)) + \cos(x)$$$⋅$$$\frac{dy}{dx} = \cos(x)$$

$$\cos(x)$$$⋅$$$\frac{dy}{dx} = \cos(x) + y\sin(x)$$

$$\frac{dy}{dx} = \frac{\cos(x) + \tan(x)\sin(x)}{\cos(x)} = 1 + \tan^2(x) = \sec^2(x)$$

arcsine(x)


$$y = \arcsin(x); \|x\| <= 1$$

$$x = \sin(y)$$

$$\frac{dx}{dy} = \cos(y)$$

$$\frac{dy}{dx} = \frac{1}{\cos(y)} = \frac{1}{\sqrt{1 - \sin ^2(y)}} = \frac{1}{\sqrt{1 - x^2}}$$

In the figure to the right you can see that the curves $$y = \arcsin(x),\ x = \sin(y)$$ are the same curve. However curve $$y = \arcsin(x)$$ is limited to $$\frac{\pi}{2} >= y >= \frac{-\pi}{2}.$$

The derivative $$\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}$$ shows that the slope of $$y = \arcsin(x)$$ is $$1$$ when $$x = 0$$ and infinite when $$x = \pm1.$$

arccosine(x)


$$y = \arccos(x); \|x\| <= 1$$

$$x = \cos(y)$$

$$\frac{dx}{dy} = -\sin(y)$$

$$\frac{dy}{dx} = \frac{-1}{\sin(y)} = \frac{-1}{\sqrt{1 - \cos ^2(y)}} = \frac{-1}{\sqrt{1 - x^2}}$$

In the figure to the right you can see that the curves $$y = \arccos(x),\ x = \cos(y)$$ are the same curve. However curve $$y = \arccos(x)$$ is limited to $$\pi >= y >= 0.$$

The derivative $$\frac{dy}{dx} = \frac{-1}{\sqrt{1 - x^2}}$$ shows that the slope of $$y = \arccos(x)$$ is $$-1$$ when $$x = 0$$ and infinite when $$x = \pm1.$$

arctan(x)


$$y = \arctan(x)$$

$$x = \tan(y)$$

$$\frac{dx}{dy} = \sec^2(y)$$

$$\frac{dy}{dx} = \frac{1}{\sec^2(y)} = \cos^2(y) = \frac{1}{1 + x^2}$$

In the figure to the right you can see that the curves $$y = \arctan(x),\ x = \tan(y)$$ are the same curve. However curve $$y = \arctan(x)$$ is limited to $$\frac{\pi}{2} > y > \frac{-\pi}{2}.$$

The derivative $$\frac{dy}{dx} = \frac{1}{\sqrt{1 + x^2}}$$ shows that the slope of $$y = \arctan(x)$$ is $$1$$ when $$x = 0$$ and $$0$$ when $$x = \pm \infty$$

a^x or $$a^x$$
$$y = a^x$$

$$y + \Delta y = a^{x + \Delta x} = a^xa^{\Delta x}$$

$$\Delta y = a^xa^{\Delta x} - a^x = a^x(a^{\Delta x} - 1)$$

$$\frac{\Delta y}{\Delta x} = a^x \cdot \frac{a^{\Delta x} - 1}{\Delta x}$$

Consider the value $$\frac{a^{\Delta x} - 1}{\Delta x}$$ specifically $$\lim_{\Delta x\rightarrow 0}\frac{a^{\Delta x} - 1}{\Delta x}$$. L'Hôpital's rule cannot be used here because $$\frac{d(a^x)}{dx}$$ is what we are trying to find.

$$a^{\frac{1}{2}} = \sqrt{a}$$

$$(a^{\frac{1}{2}})^{\frac{1}{2}} = \sqrt{\sqrt{a}}$$

$$a^{\frac{1}{2^3}} = \sqrt{\sqrt{\sqrt{a}}}$$

$$a^{\frac{1}{2^n}} = \sqrt{a}$$ taken $$n$$ times.

We will look at the expression $$\frac{a^{\Delta x} - 1}{\Delta x}$$ for different values of $$a$$ with $$\Delta x = \frac{1}{2^{70}} = 8.5e-22$$ (approx) in which case the expression becomes $$(a^{1/2^{70}} - 1)*(2^{70})$$ where $$a^{1/2^{70}}$$ is $$\sqrt[70]{a}$$ or $$\sqrt{a}$$ taken $$70$$ times. This approach is used here because function  can be written so that it does not depend on logarithmic or exponential operations.

Compare the values  with  :

We know that $$8 = 2^3;\ 32 = 2^5;\ 128 = 2^7$$. The values  are behaving like logarithms. In fact $$\lim_{\Delta x\rightarrow 0}\frac{a^{\Delta x} - 1}{\Delta x}$$ is the natural logarithm of $$a$$ written as $$\ln(a).$$



Figure 5 contains graphs of $$\frac{a^x - 1}{x}$$ for $$a = 2,\ 8,\ 32,\ 128$$ with graph of $$\frac{e^x - 1}{x}$$ included for reference.

All values of $$x$$ are valid for all curves except where $$x = 0.$$

The correct value of $$\ln(2)$$ is:

Our calculation produced:

accurate to 21 places of decimals, not bad for one line of simple python code using high-school math.

This method for calculation of $$\ln(a)$$ supposes that function  is available.

Programming language python interprets expression  as $$a^b.$$ Therefore, in python, $$\ln(2)$$ can be calculated in accordance with the basic definition above:

When $$y = a^x, \frac{dy}{dx} = a^x \cdot \ln(a).$$

The base of natural logarithms is the value of $$a$$ that gives $$\lim_{\Delta x\rightarrow 0}\frac{a^{\Delta x} - 1}{\Delta x} = 1.$$

This value of $$a$$, usually called $$e, = 2.718281828459045235360287471352662497757247093699959574.....$$

When $$a = e, \frac{d(e^x)}{dx} = e^x$$$⋅$$$\ln(e) = e^x$$$⋅$$$1 = e^x.$$

ln(x)
$$y = ln(x)$$

$$x = e^y$$

$$\frac{dx}{dy} = e^y$$

$$\frac{dy}{dx} = \frac{1}{e^y} = \frac{1}{x}$$

Examples
$$y = \ln(ax)$$

$$\frac{dy}{dx} = \frac{d}{dx}\ln(ax) = \frac{d}{dx}(\ln(a) + \ln(x)) = \frac{d}{dx}\ln(a) + \frac{d}{dx}\ln(x) = \frac{1}{x}$$

$$y = \ln(x^a)$$

$$\frac{dy}{dx} = \frac{d}{dx}\ln(x^a) = \frac{d}{dx}(a \cdot \ln(x)) = a \cdot \frac{d}{dx}\ln(x) = \frac{a}{x}$$

$$y = x^{\frac{m}{n}}$$ Calculate $$\frac{dy}{dx}$$

$$y^n = x^m$$

$$n\cdot \ln(y) = m\cdot \ln(x)$$

$$n\cdot\frac{1}{y}\cdot\frac{dy}{dx} = m\cdot \frac{1}{x}$$

$$\frac{dy}{dx} = m\cdot \frac{1}{x}\cdot\frac{y}{n}$$ $$= \frac{m}{n}\cdot\frac{x^{\frac{m}{n}}}{x}$$ $$= \frac{m}{n}\cdot x^{\frac{m}{n} - 1 }$$

Careful manipulation of logarithms converts exponents into simple constants.

Chain rule
Used where $$y = G(H(I(x)))$$

Examples
$$y = \cos(2x)$$

Let $$y = \cos(u)$$ where $$u = 2x$$.


 * $$\begin{align}

\frac{dy}{dx} =& \frac{dy}{du}\cdot\frac{du}{dx}\\ =& -\sin(u)\cdot    2\\ =& -2\sin(2x)\\ \end{align}$$

$$y = \sin^2(\ln(5x))$$

Let $$y = u^2$$ where $$u = \sin(v);\ v = \ln(w);\ w = 5x$$.


 * $$\begin{align}

\frac{dy}{dx} =& \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dw} \cdot \frac{dw}{dx}\\ =& 2u \cdot \cos(v) \cdot \frac{1}{w} \cdot 5\\ =& 2\sin(v) \cdot \cos(\ln(w)) \cdot \frac{1}{5x} \cdot 5\\ =& 2\sin(\ln(w)) \cdot \cos(\ln(5x)) \cdot \frac{1}{x}\\ =& 2\sin(\ln(5x)) \cdot \cos(\ln(5x)) \cdot \frac{1}{x}\\ \end{align}$$

Shape of curves
The first derivative of $$f(x) = y'$$ or $$f'(x).$$ As shown above, $$f'(x)$$ at any point $$x_1$$ gives the slope of $$f(x)$$ at point $$x_1.$$

$$f'(x_1)$$ is the slope of $$f(x)$$ when $$x = x_1.$$

In the example to the right, $$y = f(x) = x^2 - x - 2$$ and $$y' = f'(x) = 2x - 1.$$

Of special interest is the point at which $$f'(x)$$ or slope of $$f(x) = 0.$$

When $$f'(x) = 0,\ x = 0.5$$ and $$f(x) = 0.5^2 - 0.5 - 2 = -2.25.$$

The point $$(0.5, -2.25)$$ is called a critical point or stationary point of $$f(x).$$ Because $$y'$$ has exactly one solution for $$y' = 0,\ f(x)$$ has exactly one critical point.

The value of $$y$$ at point $$(0.5, -2.25)$$ is less than $$y$$ at both $$(0, -2),\ (1,-2).$$ Therefore the critical point $$(0.5, -2.25)$$ is a minimum of $$f(x).$$

In this curve $$y = x^2 - x - 2,$$ the point $$(0.5, -2.25)$$ is both local minimum and absolute minimum.

In the example to the right, $$y = f(x) = \frac{2x^3 + 3x^2 - 12x - 8}{8}$$ and $$y' = f'(x) = \frac{6x^2 + 6x - 12}{8}.$$

Of special interest are the points at which $$f'(x)$$ or slope of $$f(x) = 0.$$

When $$f'(x) = 0,\ x = -2$$ and $$f(x) = 1.5$$ or

When $$f'(x) = 0,\ x = 1$$ and $$f(x) = -1\frac{7}{8}.$$

The points $$(-2, 1.5),\ (1, -1\frac{7}{8})$$ are critical or stationary points of $$f(x).$$ Because $$y'$$ has exactly two real solutions for $$y' = 0,\ f(x)$$ has exactly two critical points.

Slope of $$f(x)$$ to the left of $$(-2, 1.5)$$ is positive and adjacent slope of $$f(x)$$ to the right of $$(-2, 1.5)$$ is negative. Therefore point $$(-2, 1.5)$$ is local maximum. Point $$(-2, 1.5)$$ is not absolute maximum.

Adjacent slope of $$f(x)$$ to the left of $$(1, -1\frac{7}{8})$$ is negative and slope of $$f(x)$$ to the right of $$(1, -1\frac{7}{8})$$ is positive. Therefore point $$(1, -1\frac{7}{8})$$ is local minimum. Point $$(1, -1\frac{7}{8})$$ is not absolute minimum.

Electric water heater
A cylindrical water heater is standing on its base on a hard rubber pad that is a perfect thermal insulator. The vertical curved surface and the top are exposed to the free air. The design of the cylinder requires that the volume of the cylinder should be maximum for a given surface area exposed to the free air. What is the shape of the cylinder?

Let the height of the cylinder be $$h$$ and let $$h = Kr$$ where $$r$$ is the radius and $$K$$ is a constant.

Surface of cylinder $$ = S = \pi r^2 + 2\pi r h = \pi r^2 + 2\pi r (Kr) = \pi r^2 + 2\pi r^2K$$

Volume of cylinder $$ = V = \pi r^2h = \pi r^2(Kr) = \pi r^3K$$

$$K = \frac{S - \pi r^2}{2\pi r^2}$$

$$V = \pi r^3 \cdot \frac{S - \pi r^2}{2\pi r^2} = \frac{r(S - \pi r^2)}{2} = \frac{rS - \pi r^3}{2}$$

$$\frac{dV}{dr} = \frac{S - 3\pi r^2}{2}$$

For maximum volume, $$\frac{dV}{dr} = 0$$

Therefore $$S - 3\pi r^2 = 0$$

$$S = 3\pi r^2;\ \pi r^2(1 + 2K) = 3\pi r^2;\ 1 + 2K = 3;\ K =1.$$

The height of the cylinder equals the radius.

Square and triangle
A square of side $$h$$ has perimeter $$P_s = 4h$$ and area $$A_s = h^2.$$

An equilateral triangle of side $$s$$ has perimeter $$P_t = 3s$$ and area $$A_t = \frac{\sqrt{3}}{4}s^2.$$

$$P_s + P_t = 50.$$

Total area $$A = A_s + A_ t$$ and $$A$$ must be minimum. What is the value of $$s$$?

$$4h + 3s = 50$$

$$h = \frac{50 - 3s}{4}$$

$$A_s = h^2 = (\frac{50 - 3s}{4})^2$$ $$= \frac{2500 - 300s + 9s^2}{16}$$

$$A = A_s + A_t$$ $$= \frac{2500 - 300s + 9s^2}{16} + \frac{\sqrt{3}}{4}s^2$$ $$= \frac{1}{16}(2500 - 300s + 9s^2 + 4\sqrt{3}s^2)$$

For minimum $$A,\ \frac{dA}{ds} = 0.$$

$$-300 + 2(9 + 4\sqrt{3})s = 0$$

$$(9 + 4\sqrt{3})s = 150$$

$$s = \frac{150}{9 + 4\sqrt{3}}$$ $$= 9.417\dots$$

County Road


Town B is 40 miles East and 50 miles North of Town A. The county is going to construct a road from Town A to Town B. Adjacent to Town A the cost to build a road is $500k/mile. Adjacent to Town B the cost to build a road is $200k/mile. The dividing line runs East-West 30 miles North of Town A. Calculate the position of point C so that the cost of the road from Town A to Town B is minimum.

Let point $$C = (x,30).$$

Then distance from Town A to point $$C = \sqrt{x^2 + 30^2} = \sqrt{x^2 + 900}.$$

Distance from Town B to point $$C = \sqrt{(40-x)^2 + 20^2} $$ $$= \sqrt{1600 - 80x + x^2 + 400} $$ $$= \sqrt{2000 - 80x + x^2}.$$



Cost $$= 5\sqrt{x^2 + 900} + 2\sqrt{2000 - 80x + x^2}$$ in units of $100k.

For minimum cost $$f'(x) = 0.$$

$$5\cdot\frac{1}{2}\cdot\frac{2x}{\sqrt{x^2 + 900}} + 2\cdot\frac{1}{2}\cdot\frac{-80+2x}{\sqrt{2000-80x+x^2}} = 0$$

$$\frac{5x}{\sqrt{x^2+900}} + \frac{-80+2x}{\sqrt{2000-80x+x^2}} = 0$$

$$5x\sqrt{2000-80x+x^2} + (-80+2x)\sqrt{x^2+900} = 0$$

$$x = 10.52322823517\dots$$

Cardboard box


A piece of cardboard of length $$4\ ft$$ and width $$3\ ft$$ will be used to make a box with a top. Some waste will be cut out of the piece of cardboard and the remaining cardboard will be folded to make a box so that the volume of the box is maximum.

What is the height of the box?



$$2h + l = 3$$

$$2w + 2h = 4;\ w + h = 2$$

$$V = lwh = (3 - 2h)(2 - h)h = 2h^3 - 7h^2 + 6h$$

For maximum volume $$\frac{dV}{dh} = 6h^2 - 14h + 6 = 0.$$

$$3h^2 - 7h + 3 = 0$$

$$h = \frac{7 - \sqrt{13}}{6} = 6.788\dots$$ inches.

Solving ellipse at origin


An ellipse with center at origin has equation: $$Ax^2 + Bxy + Cy^2 + F = 0\ \dots\ (1)$$

Given values $$A, B, C, F$$ calculate:


 * length of major axis


 * length of minor axis.

In Figure 1 $$t$$ is any line from origin to ellipse and $$\theta$$ is angle between $$X$$ axis and $$t.$$

Aim of this section is to calculate $$\theta$$ so that length of $$t$$ is maximum, in which case length of major axis = $$2t.$$

Let $$c = \cos(\theta)$$ and $$s = \sin(\theta).$$

Then $$x = t \cos(\theta) = tc,$$ and $$y = t \sin(\theta) = ts.$$

Substitute these values in $$(1):$$

$$Attcc + Bttcs + Cttss + F = 0\ \dots\ (2)$$

Calculate $$t' = \frac{dt}{d\theta}$$

$$A(tt(-2cs) + cc2tt') + B(tt(cc-ss) + cs2tt') + C(tt(2sc) + ss2tt') = 0$$

$$Att(-2cs) + Acc2tt' + Btt(cc-ss) + Bcs2tt' + Ctt(2sc) + Css2tt' = 0$$

$$-2Attcs + Acc2tt' + Bttcc-Bttss + Bcs2tt' + 2Cttsc + 2Csstt' = 0$$

$$ + Acc2tt' + 2Bcstt' + 2Csstt' - 2Attcs + Bttcc - Bttss + 2Cttsc = 0$$

$$ + t'(Acc2t + 2Bcst + 2Csst) = + 2Attcs - Bttcc + Bttss - 2Cttsc$$

$$t' = \frac{+ 2Attcs - Bttcc + Bttss - 2Cttsc}{(Acc2t + 2Bcst + 2Csst)}$$

For maximum or minimum $$t:$$

$$2Attcs - Bttcc + Bttss - 2Cttsc = 0$$

$$2Acs - Bcc + Bss - 2Csc = 0$$

$$2Acs - 2Csc = Bcc - Bss $$

Square both sides, substitute $$1 - ss$$ for $$cc$$ and result is:

$$aS^2 + bS + c = 0\ \dots\ (3)$$ where:

$$S = \sin^2(\theta)$$

$$a = (+ 4AA - 8AC + 4BB + 4CC)$$

$$b = -a$$

$$c = BB$$

An example
Let equation of ellipse be: $$55x^2 - 24xy + 48y^2 - 2496 = 0$$ The solutions of quadratic equation $$(3)$$ are $$.36$$ or $$.64$$.

Therefore $$\sin(\theta) = \pm 0.6$$ or $$\sin(\theta) = \pm 0.8$$.

From $$(2)$$ above: $$t = \sqrt{\frac{-F}{Acc + Bcs + Css}}$$

Minimum value of $$t = 6.244997998398398.$$ Length of minor axis $$= 2 * 6.244997998398398$$

Maximum value of $$t = 7.999999999999999.$$ Length of major axis $$= 8 * 2$$

The car jack


In triangle $$ABC$$ to the right:


 * length $$AB = 10$$ inches,


 * length $$BC = x$$ inches and is horizontal,


 * length $$CA = y$$ inches and is vertical,


 * $$\angle ABC = \theta.$$

Point $$B$$ is moving towards point $$C$$ at the rate of $$5$$ inches$$/$$minute.

Vertical motion
[[File:0216_0901CarJack3.png|thumb|400px| Figure 3: Curves and values associated with car jack.

When $$x = 1,\ \frac{dy}{dx} = 0.1\dots$$ When $$\theta = 45^\circ,\ \frac{dy}{dx} = 1$$ When $$x = 9,\ \frac{dy}{dx} = 2.06\dots$$ ]] At what rate is point $$A$$ moving upwards:

(a) when $$x = 9$$?

(b) when $$x = 1$$?

(c) when $$\theta = 45^\circ$$?

We have to calculate $$\frac{dy}{dt}$$ when $$\frac{dx}{dt}$$ is given.

$$\frac{dy}{dt} = \frac{dy}{dx}\cdot\frac{dx}{dt}$$

$$x^2 + y^2 = 10^2$$ (equation of circle)

$$y = \sqrt{100 - x^2}$$

$$y^2 = 100 - x^2$$

$$2y\cdot\frac{dy}{dx} = -2x$$

$$\frac{dy}{dx} = \frac{-2x}{2y} = \frac{-x}{\sqrt{100 - x^2}}$$

For convenience we'll use the negative value of the square root and say that $$\frac{dy}{dx} = \frac{x}{\sqrt{100 - x^2}}.$$

Relative to line $$BC:$$

When $$x = 9,\ \frac{dy}{dx} = 2.06,\ \frac{dy}{dt} = 2.06\cdot5 = 10.3$$ inches$$/$$minute.

When $$x = 1,\ \frac{dy}{dx} = 0.1,\ \frac{dy}{dt} = 0.1\cdot5 = 0.5$$ inches$$/$$minute.

When $$\theta = 45^\circ, x = 10\cdot\cos 45^\circ = 7.071$$ and $$\frac{dy}{dt} = 1\cdot 5 = 5$$ inches$$/$$minute.

This example highlights the mechanical advantage of this simple but effective tool. When the top of the jack is low, it moves quickly. As the jack takes more and more weight, the top of the jack moves more slowly.

Change of area


At what rate is the area of $$\Delta ABC$$ changing when

(i) $$x = 9$$?

(ii) $$x = 1$$?

(iii) $$\theta = 45^{\circ}$$?

$$\frac{da}{dt} = \frac{da}{dx}\cdot\frac{dx}{dt}$$ where $$a$$ is area of $$\Delta ABC$$ and $$\frac{dx}{dt} = 5$$ inches $$/$$ minute.

$$a = \frac{b\cdot h}{2} = \frac{x \cdot y}{2} = \frac{x\sqrt{100 - x^2}}{2}$$

Calculate $$\frac{da}{dx}:$$

$$2a = x\sqrt{100 - x^2}$$

$$4a^2 = x^2(100 - x^2) = 100x^2 - x^4$$

$$8a\cdot\frac{da}{dx} = 200x - 4x^3$$

$$\frac{da}{dx} = \frac{200x - 4x^3}{8a} = \frac{50x - x^3}{2a} = \frac{50x - x^3}{ x\sqrt{100 - x^2} }$$ $$ = \frac{50 - x^2}{ \sqrt{100 - x^2} }$$

When $$x = 9,\ \frac{da}{dx} = 7.11\dots$$ and area of $$\Delta ABC$$ is increasing at rate of $$(5\cdot 7.11\dots)$$ square inches/minute.

When $$x = 1,\ \frac{da}{dx} = -4.92\dots$$ and area of $$\Delta ABC$$ is decreasing at rate of $$(5\cdot 4.92\dots)$$ square inches/minute.

When $$\theta = 45^{\circ},\ h = b = x = y = 10\cos(45^{\circ}) = 10\cdot\frac{\sqrt{2}}{2} = 5\sqrt{2},\ $$ $$\frac{da}{dx} = 0,\ a$$ is a maximum of $$25$$ square inches and $$\frac{da}{dt} = 0.$$ $$$$

On the clock
An old fashion analog clock with American style face (12 hours) keeps accurate time. The length of the minute hand is $$4$$ inches and the length of the hour hand is $$3$$ inches.

At what rate is the tip of the minute hand approaching the tip of the hour hand at 3 o'clock?

Let $$a$$ be distance between the two tips.

The task is to calculate $$\frac{da}{dt}.$$

$$\frac{da}{dt} = \frac{da}{d\theta}\cdot \frac{d\theta}{dt}$$ where $$\theta$$ is angle subtended by side $$a$$ at center of clock.

Calculating $$\frac{d\theta}{dt}:$$

Angular velocity of minute hand $$= \frac{2\pi}{1}$$ radians/hour.

Angular velocity of hour hand $$= \frac{2\pi}{12} = \frac{\pi}{6}$$ radians/hour.

$$\frac{d\theta}{dt} =$$ angular velocity of minute hand relative to hour hand $$= 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}$$ radians/hour.

Calculating $$\frac{da}{d\theta}:$$

$$a^2 = 3^2 + 4^2 - 2\cdot 3\cdot 4\cdot\cos \theta = 25 - 24\cos \theta.$$

$$2a\cdot\frac{da}{d\theta} = 24\sin \theta.$$

$$\frac{da}{d\theta} = \frac{24\sin \theta}{2a} = \frac{24\sin 90^\circ}{2\cdot 5} = \frac{24}{10} = 2.4.$$

$$\frac{da}{dt} = 2.4\cdot\frac{11\pi}{6} = 4.4\pi$$ inches/hour.

Radar Speed Trap


A multi-lane highway is oriented East-West. A vehicle is moving in the inside lane from West to East. A law-enforcement officer with a radar gun is in position 50 feet South of the center of the inside lane. When the vehicle is 200 feet from the radar gun, it shows the vehicle's speed to be 60 mph. What is the actual speed of the vehicle?

Let length $$AC = b.$$

$$\frac{db}{dt} = \frac{db}{dc}\cdot\frac{dc}{dt}$$ where $$\frac{dc}{dt} = 60$$ mph.

The derivative $$\frac{db}{dc}:$$
 * $$\begin{align}

\ \ \ \ &b^2 = c^2 - a^2;\ b = \sqrt{c^2 - a^2}\\ &2b\cdot \frac{db}{dc} = 2c\\ &\frac{db}{dc} = \frac{2c}{2b} = \frac{c}{\sqrt{c^2 - a^2}} \end{align}$$

$$\frac{db}{dt} = \frac{200}{\sqrt{200^2 - 50^2}}\cdot 60 = 61.967\dots$$ mph.

On the Water
A ship is at sea $$15$$ nautical miles East and $$8$$ nautical miles South of a lighthouse, and the ship is steaming South-West at $$10$$ nautical miles/hour or $$10$$ knots.

At what rate is ship approaching lighthouse?

$$\frac{da}{dt} = 10$$ knots.

$$\frac{dc}{dt} = \frac{dc}{da}\cdot\frac{da}{dt}$$ knots.

Calculating $$\frac{dc}{da}:$$

$$c^2 = a^2 + b^2 - 2ab\cos\theta$$

$$2c\cdot\frac{dc}{da} = 2a - 2b\cos\theta$$

$$\frac{dc}{da} = \frac{a - b\cos\theta}{c} = -0.291161615782696$$

The $$-$$ sign indicates that, when $$a$$ is increasing, $$c$$ is decreasing.

$$\frac{dc}{dt} = \frac{dc}{da}\cdot10$$ $$= 2.91161615782696$$ knots towards the lighthouse.

Reciprocating engine
The diagram illustrates the piston, connecting rod and crankshaft of an internal combustion reciprocating engine.

The connecting rod $$a$$ is connected to piston on $$y$$ axis, and to crankshaft at end of $$c$$ on $$x$$ axis.

This section analyzes the motion of the piston as the crankshaft rotates through angle $$x$$ and the piston moves up and down on the $$y$$ axis.

Position of piston
$$a^2 = y^2 + c^2 - 2yc\cos(x)$$

$$y^2 - 2c\cos(x)y + c^2 - a^2 = 0$$

$$y^2 - 2(5)\cos(x)y + (5)^2 - (13)^2 = 0$$

$$y^2 - 10\cos(x)y - 144 = 0$$

$$y = \frac{10\cos(x) + \sqrt{ (10\cos(x))^2 -4(-144) } }{2}$$

$$y = \frac{10\cos(x) + \sqrt{ 100(\cos(x))^2 + 576 } }{2}$$

$$y = 5\cos(x) + \sqrt{ 25(\cos(x))^2 + 144 }$$

Code supplied to grapher (without white space) is: When expressed in this way, it's easy to convert the code to python code:

Positions of interest:

Velocity of piston
Calculation of $$\frac{dy}{dx}$$ by implicit differentiation.

$$y^2 - 10 \cos(x)y - 144 = 0$$

$$2y\cdot y' - 10(\cos(x)y' + y(-\sin(x)) ) = 0$$

$$2y\cdot y' - 10\cos(x)y' + 10\sin(x)y = 0$$

$$y\cdot y' - 5\cos(x)y' = -5\sin(x)y$$

$$y'(y - 5\cos(x)) = -5\sin(x)y$$

$$y' = \frac{5\sin(x)y}{5\cos(x) - y}$$

Acceleration of piston
Acceleration introduces the second derivative. While velocity was the first derivative of position with respect to time, acceleration is the first derivative of velocity or the second derivative of position.

From velocity above $$yy' - 5\cos(x)y' + 5\sin(x)y = 0$$

By implicit differentation:

$$yy + y'y' - 5(\cos(x)y + y'(-\sin(x))) + 5(\sin(x)y' + y\cos(x)) = 0$$

$$yy + y'y' - 5\cos(x)y + 5y'\sin(x) + 5\sin(x)y' + 5y\cos(x) = 0$$

$$yy - 5\cos(x)y = -( y'y' + 5y'\sin(x) + 5\sin(x)y' + 5y\cos(x) )$$

$$y''(y - 5\cos(x)) = -( y'y' + 10\sin(x)y' + 5y\cos(x) )$$

$$y'' = \frac{y'y' + 10\sin(x)y' + 5y\cos(x) }{5\cos(x) - y}$$

Substitute for $$y$$ and $$y'$$ as defined above, and you see the code input to grapher at top of diagram to right.

"Kinks" in the curve:

It is not obvious by looking at the curve of velocity that there are slight irregularities in the curve when velocity is increasing.

However, the irregularities are obvious in the curve of acceleration.

During one revolution of the crankshaft there is less time allocated for negative acceleration than for positive acceleration. Therefore, the maximum absolute value of negative acceleration is greater than the maximum value of positive acceleration.

Minimum and maximum velocity
Velocity is rate of change of position. See also Figure 3 above.

Minimum velocity:

Velocity is zero when slope of curve of position is zero. This occurs at top dead center and at bottom dead center, ie, when $$x = 0$$ and $$x = \pi.$$

Maximum velocity:

Intuition suggests that the position of maximum velocity might be the point at which the connecting rod is tangent to the circle of the crankshaft. In other words:

$$x = \arctan(\frac{13}{5}) = 68.96\dots^{\circ}$$

or, that the position of maximum velocity might be the point at which the piston is half-way between top dead center and bottom dead center. In other words:

$$x = \arccos(\frac{2.5}{13}) = 78.9\dots^{\circ}$$

However, velocity is maximum when acceleration is $$0,$$ which occurs when $$x = \pm 71.26\dots^{\circ}.$$

Suppose that the engine is rotating at $$100$$ radians/second or approx. $$955$$ RPM.

$$v_{max} = y' \frac{inches}{radian} \cdot 100\frac{radians}{second}$$

abs$$(v_{max}) = 5.36 (100)$$ inches/second or approx. $$30.5$$ mph.

Minimum and maximum acceleration
Acceleration is rate of change of velocity. See also Figures 4a and 4b above.

Minimum acceleration:

Acceleration is zero when slope of curve of velocity is zero. This occurs at maximum velocity or when $$\angle x$$ is approx. $$\pm 71.3^{\circ}.$$

Maximum acceleration:

Acceleration is maximum when slope of curve of velocity is maximum.

Maximum negative acceleration occurs when slope of curve of velocity is maximum negative. This happens at top dead center when $$\angle x = 0.$$

Maximum positive acceleration occurs when slope of curve of velocity is maximum positive. This happens before and after bottom dead center when $$\angle x$$ is approx. $$\pm 127^{\circ}.$$

Let the engine continue to rotate at $$100$$ radians/second.

$$acc_{max} = y''\frac{inches}{radian^2}\cdot 100\frac{radians}{second} \cdot 100\frac{radians}{second}$$

abs$$(acc_{max}) = 6.92*100*100$$ inches/second$$^2$$ or approx. $$180$$ times the acceleration due to terrestrial gravity.

This maximum value of acceleration is maximum negative when $$\angle x = 0.$$

According to Newtonian physics $$f = ma$$, force = mass*acceleration, and $$w = fs$$, work = force*distance. In this engine energy expended in just accelerating piston to maximum velocity is proportional to rpm$$^2$$.

Perhaps this helps to explain why a big marine diesel engine rotating at low RPM can achieve efficiency of $$55 \%.$$

Simple laws of motion
Let a body move in accordance with the following function of $$t,\ f(t)$$ where $$t$$ means time:

$$p_t = f(t) = at^2 + bt + c$$ where $$p_t$$ is position at time $$t.$$

$$p_t$$ has the dimension of length. Therefore, each component of $$f(t)$$ must have the dimension of length.

For $$bt$$ to have the dimension of length, $$b$$ must have the dimensions of $$\frac{length}{time}$$ or velocity.

For $$at^2$$ to have the dimension of length, $$a$$ must have the dimensions of $$\frac{length}{time^2}$$ or acceleration.

If $$t ==0,\ p_0 = c$$ and $$p_t = f(t) = at^2 + bt + p_0.$$

The derivatives enable us to assign specific values to $$a,\ b.$$

$$\frac{d(p_t)}{dt} = f'(t) = v_t = 2at + b$$ where $$v_t$$ is velocity at time $$t.$$

If $$t ==0,\ v_0 = b$$ and $$p_t = f(t) = at^2 + v_0t + p_0.$$

$$\frac{d(v_t)}{dt} = f''(t) = 2a,$$ a constant equal to the acceleration to which the body is subjected.

For convenience let us say that $$p_t = \frac{1}{2}at^2 + v_0t + p_0$$ where $$a$$ is the (constant) acceleration to which the body is subjected.

Then $$v_t = \frac{1}{2}\cdot 2at + v_0 = at + v_0$$ and $$\frac{d(v_t)}{dt} = a.$$