maths > differential-calculus

Understanding Algebra of Derivatives

what you'll learn...

Overview

»  Understanding Algebra of Derivatives
how derivative applies to a function given as algebraic operations of several functions

→  product and division

→  function-of-function

→  parametric form of function

»  Differentiation under Basic Arithmetic Operations

→  $\left(au\right)\prime =au\prime$

→  $\left(u+v\right)\prime =u\prime +v\prime$

→  $\left(u-v\right)\prime =u\prime -v\prime$

→  $\left(uv\right)\prime =u\prime v+uv\prime$

→  $\left(\frac{u}{v}\right)\prime =\frac{u\prime v-uv\prime }{{v}^{2}}$

»  Differentiation under Function Operations

→  Composite form and Chain rule: given $v\left(u\right)$$v \left(u\right)$ and $u\left(x\right)$$u \left(x\right)$ (i.e. $v\left(u\left(x\right)\right)$$v \left(u \left(x\right)\right)$) then
$\frac{dv}{dx}=\frac{dv}{du}\frac{du}{dx}$$\frac{\mathrm{dv}}{\mathrm{dx}} = \frac{\mathrm{dv}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$

→  Parametric form : given $g\left(x\right)$$g \left(x\right)$ and $h\left(x\right)$$h \left(x\right)$ then
$\frac{dg}{dh}=\frac{dg/dx}{dh/dx}$$\frac{\mathrm{dg}}{\mathrm{dh}} = \frac{\mathrm{dg} / \mathrm{dx}}{\mathrm{dh} / \mathrm{dx}}$

algebra is about operations + - * / ^

"Algebra of differentiation" or "Algebra of derivatives" means studying of "Properties to find derivatives of functions given as algebraic operations of several functions".

The mathematical operations with functions $u\left(x\right)$$u \left(x\right)$ and $v\left(x\right)$$v \left(x\right)$ are

•  addition and subtraction $u\left(x\right)±v\left(x\right)$$u \left(x\right) \pm v \left(x\right)$

•  multiple of a function $au\left(x\right)$$a u \left(x\right)$

•  multiplication and division $u\left(x\right)v\left(x\right)$$u \left(x\right) v \left(x\right)$ and $\frac{u\left(x\right)}{v\left(x\right)}$$\frac{u \left(x\right)}{v \left(x\right)}$

•  powers and roots ${\left[u\left(x\right)\right]}^{n}$${\left[u \left(x\right)\right]}^{n}$ and ${\left[u\left(x\right)\right]}^{\frac{1}{n}}$${\left[u \left(x\right)\right]}^{\frac{1}{n}}$

•  composite form of functions $v\left(u\left(x\right)\right)$$v \left(u \left(x\right)\right)$

•  parametric form of functions $v=f\left(r\right);u=g\left(r\right)$

establishing the problem

Given that $f\left(x\right)=u\left(x\right)\star v\left(x\right)$$f \left(x\right) = u \left(x\right) \star v \left(x\right)$ where $\star$$\star$ is one of the arithmetic or function operations.

Will there be any relationship between the derivative of the functions $\frac{d}{dx}u\left(x\right)$$\frac{d}{\mathrm{dx}} u \left(x\right)$ ; $\frac{d}{dx}v\left(x\right)$$\frac{d}{\mathrm{dx}} v \left(x\right)$ and the derivative of the result $\frac{d}{dx}f\left(x\right)$$\frac{d}{\mathrm{dx}} f \left(x\right)$?

Algebra of differentiation analyses this and provides the required knowledge.

Note: In deriving the results, the functions are assumed to be continuous and differentiable at the points or range of interest. For specific functions at specific values of variables, one must check for the continuity and the differentiability before using the algebra of derivatives.

example problem

For example, consider
$u\left(x\right)={x}^{2}$$u \left(x\right) = {x}^{2}$
$v\left(x\right)=\mathrm{sin}x$$v \left(x\right) = \sin x$
$f\left(x\right)={x}^{2}\mathrm{sin}x$$f \left(x\right) = {x}^{2} \sin x$

From the standard results, it is known that
$\frac{d}{dx}{x}^{2}=2x$$\frac{d}{\mathrm{dx}} {x}^{2} = 2 x$ and
$\frac{d}{dx}\mathrm{sin}x=\mathrm{cos}x$$\frac{d}{\mathrm{dx}} \sin x = \cos x$.
What is $\frac{d}{dx}{x}^{2}\mathrm{sin}x$$\frac{d}{\mathrm{dx}} {x}^{2} \sin x$?

In this particular example multiplication is considered. Instead of multiplication, one of the arithmetic or function operations may be considered too.

The algebra of derivatives analyses this and provides the required knowledge.

scalar multiple

Derivative of a scalar multiple of function. Given $v\left(x\right)=au\left(x\right)$$v \left(x\right) = \textcolor{\mathrm{de} e p s k y b l u e}{a} u \left(x\right)$.

$\underset{\delta \to 0}{lim}\frac{v\left(x+\delta \right)-v\left(x\right)}{\delta }$${\lim}_{\delta \to 0} \frac{v \left(x + \delta\right) - v \left(x\right)}{\delta}$

$\quad \quad = {\lim}_{\delta \to 0} \frac{\textcolor{\mathrm{de} e p s k y b l u e}{a} u \left(x + \delta\right) - \textcolor{\mathrm{de} e p s k y b l u e}{a} u \left(x\right)}{\delta}$

with continuity and differentiability conditions on $u$$u$ and $a$$a$
$\quad \quad = \textcolor{\mathrm{de} e p s k y b l u e}{a} {\lim}_{\delta \to 0} \frac{u \left(x + \delta\right) - u \left(x\right)}{\delta}$

$\quad \quad = \textcolor{\mathrm{de} e p s k y b l u e}{a} \frac{d}{\mathrm{dx}} u$

The above proves that $\left(au\right)\prime =au\prime$. In other words, derivative of a multiple of a function is multiple of the derivative of the function

Intuitive understanding for
$\left(au\right)\prime =au\prime$
rate of change multiplies when the function is multiplied by a constant.

Given $\frac{dy}{dx}=2{x}^{2}$$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 {x}^{2}$ and $v=\frac{y}{5}$$v = \frac{y}{5}$, what is $\frac{dv}{dx}$$\frac{\mathrm{dv}}{\mathrm{dx}}$?

The answer is "$\frac{2}{5}{x}^{2}$$\frac{2}{5} {x}^{2}$"

$\frac{dv}{dx}$$\frac{\mathrm{dv}}{\mathrm{dx}}$
$=\frac{d}{dx}\frac{y}{5}$$= \frac{d}{\mathrm{dx}} \frac{y}{5}$
$=\frac{1}{5}\frac{d}{dx}y$$= \frac{1}{5} \frac{d}{\mathrm{dx}} y$
$=\frac{1}{5}×2{x}^{2}$$= \frac{1}{5} \times 2 {x}^{2}$
$=\frac{2}{5}{x}^{2}$$= \frac{2}{5} {x}^{2}$

summary

Derivative of a Multiple:
$\left(au\right)\prime =au\prime$
Derivative of a multiple of a function is multiple of the derivative of the function.

sum or difference

Finding derivative of sum or difference.

$\underset{\delta \to 0}{lim}\left[\left(u\left(x+\delta \right)±v\left(x+\delta \right)\right)$$-\left(u\left(x\right)±v\left(x\right)\right)\right]/\delta$

$±\left(v\left(x+\delta \right)-v\left(x\right)\right)\right]/\delta$

with continuity and differentiability conditions on $u$$u$ and $v$$v$
$\quad = {\lim}_{\delta \to 0} \frac{u \left(x + \delta\right) - u \left(x\right)}{\delta}$

$±\underset{\delta \to 0}{lim}\frac{v\left(x+\delta \right)-v\left(x\right)}{\delta }$$\pm {\lim}_{\delta \to 0} \frac{v \left(x + \delta\right) - v \left(x\right)}{\delta}$

$\quad \quad = \frac{d}{\mathrm{dx}} u \pm \frac{d}{\mathrm{dx}} v$

The above proves $\left(u+v\right)\prime =u\prime +v\prime$ and $\left(u-v\right)\prime =u\prime -v\prime$. In other words, Derivative of a sum or difference is the sum or difference of derivatives.

Intuitive understanding for
$\left(u±v\right)\prime =u\prime ±v\prime$

Given $\frac{dy}{dx}=\mathrm{sin}x$$\frac{\mathrm{dy}}{\mathrm{dx}} = \sin x$ and $v=y+20$$v = y + 20$, what is $\frac{dv}{dx}$$\frac{\mathrm{dv}}{\mathrm{dx}}$?

The answer is "$\mathrm{sin}x$$\sin x$"

$\frac{dv}{dx}$$\frac{\mathrm{dv}}{\mathrm{dx}}$
$=\frac{d\left(y+20\right)}{dx}$$= \frac{d \left(y + 20\right)}{\mathrm{dx}}$
$=\frac{dy}{dx}+\frac{d\left(20\right)}{dx}$$= \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{d \left(20\right)}{\mathrm{dx}}$
$=\mathrm{sin}x+0$$= \sin x + 0$

summary

Derivative of Sum or Difference:
$\left(u+v\right)\prime =u\prime +v\prime$

$\left(u-v\right)\prime =u\prime -v\prime$
Derivative of a sum or difference is the sum or difference of derivatives.

product

Finding derivative of product.

$\underset{\delta \to 0}{lim}\left[\left(u\left(x+\delta \right)×v\left(x+\delta \right)\right)$$-\left(u\left(x\right)×v\left(x\right)\right)\right]/\delta$

adding $u\left(x\right)×v\left(x+\delta \right)$$\textcolor{c \mathmr{and} a l}{u \left(x\right) \times v \left(x + \delta\right)}$$-u\left(x\right)×v\left(x+\delta \right)$$- \textcolor{\mathrm{de} e p s k y b l u e}{u \left(x\right) \times v \left(x + \delta\right)}$

$-u\left(x\right)×v\left(x+\delta \right)$$- \textcolor{\mathrm{de} e p s k y b l u e}{u \left(x\right) \times v \left(x + \delta\right)}$$+u\left(x\right)×v\left(x+\delta \right)$$+ \textcolor{c \mathmr{and} a l}{u \left(x\right) \times v \left(x + \delta\right)}$$-u\left(x\right)×v\left(x\right)\right]/\delta$

with continuity and differentiability conditions on $u$$u$ and $v$$v$
$-u\left(x\right)\right)\right]/\delta$

$-v\left(x\right)\right)\right)\right]/\delta$

$\quad \quad = v \frac{d}{\mathrm{dx}} u + u \frac{d}{\mathrm{dx}} v$

The above proves that $\left(uv\right)\prime =u\prime v+uv\prime$

Intuitive understanding of
$\left(uv\right)\prime =u\prime v+uv\prime$

Two functions are multiplied at every point. Rate of change of the product equals sum of

fixing one function $v$$v$ as a constant and rate of change of the other function $u\prime$

plus fixing the other function $u$$u$ as a constant and rate of change of the function $v\prime$

This is compared to $\left(au\right)\prime =au\prime$.

Given $y={x}^{2}\mathrm{sin}x$$y = {x}^{2} \sin x$ what is $\frac{dy}{dx}$$\frac{\mathrm{dy}}{\mathrm{dx}}$?
Note: $\frac{d}{dx}{x}^{2}=2x$$\frac{d}{\mathrm{dx}} {x}^{2} = 2 x$
$\frac{d}{dx}\mathrm{sin}x=\mathrm{cos}x$$\frac{d}{\mathrm{dx}} \sin x = \cos x$

The answer is "$2x\mathrm{sin}x+{x}^{2}\mathrm{cos}x$$2 x \sin x + {x}^{2} \cos x$"

$\frac{dy}{dx}$$\frac{\mathrm{dy}}{\mathrm{dx}}$
$=\frac{d}{dx}\left({x}^{2}\mathrm{sin}x\right)$$= \frac{d}{\mathrm{dx}} \left({x}^{2} \sin x\right)$
applying product law of derivative $=\mathrm{sin}x\frac{d}{dx}{x}^{2}+{x}^{2}\frac{d}{dx}\mathrm{sin}x$$= \sin x \frac{d}{\mathrm{dx}} {x}^{2} + {x}^{2} \frac{d}{\mathrm{dx}} \sin x$
$=2x\mathrm{sin}x+{x}^{2}\mathrm{cos}x$$= 2 x \sin x + {x}^{2} \cos x$

summary

Derivative of Product: $\left(uv\right)\prime =u\prime v+uv\prime$
Derivative of product of two function is sum of derivatives of the functions scaled by the value of other function

division

Finding derivative of division.
$\underset{\delta \to 0}{lim}\left[$$u\left(x+\delta \right)÷v\left(x+\delta \right)$$u \left(x + \delta\right) \div v \left(x + \delta\right)$$-\left(u\left(x\right)÷v\left(x\right)\right)\right]/\delta$

$=\underset{\delta \to 0}{lim}\left\{$$\frac{1}{v\left(x+\delta \right)v\left(x\right)}$$\frac{1}{v \left(x + \delta\right) v \left(x\right)}$$×\left[u\left(x+\delta \right)v\left(x\right)$$-u\left(x\right)v\left(x+\delta \right)\right]/\delta \right\}$

adding $u\left(x\right)v\left(x\right)-u\left(x\right)v\left(x\right)$$\textcolor{c \mathmr{and} a l}{u \left(x\right) v \left(x\right)} - \textcolor{\mathrm{de} e p s k y b l u e}{u \left(x\right) v \left(x\right)}$

$=\underset{\delta \to 0}{lim}\left\{$$\frac{1}{v\left(x+\delta \right)v\left(x\right)}$$\frac{1}{v \left(x + \delta\right) v \left(x\right)}$$\left[u\left(x+\delta \right)v\left(x\right)-u\left(x\right)v\left(x\right)$$-u\left(x\right)v\left(x+\delta \right)+u\left(x\right)v\left(x\right)\right]/\delta \right\}$

$=\underset{\delta \to 0}{lim}\left\{$$\frac{1}{v\left(x+\delta \right)v\left(x\right)}$$\frac{1}{v \left(x + \delta\right) v \left(x\right)}$$\left[v\left(x\right)\left(u\left(x+\delta \right)-u\left(x\right)\right)$$-u\left(x\right)\left(v\left(x+\delta \right)-v\left(x\right)\right)\right]/\delta \right\}$

$=\underset{\delta \to 0}{lim}\left\{$$\frac{1}{v\left(x+\delta \right)v\left(x\right)}×$$\frac{1}{v \left(x + \delta\right) v \left(x\right)} \times$$\left[v\left(x\right)\left(u\left(x+\delta \right)-u\left(x\right)\right)\right]/\delta \right)$$-\left[u\left(x\right)\left(v\left(x+\delta \right)-v\left(x\right)\right)\right]/\delta \right\}$

with continuity and differentiability conditions on $u$$u$ and $v$$v$
$\frac{1}{{v}^{2}}$$\frac{1}{{v}^{2}}$$×\left\{\frac{du}{dx}v$$-u\frac{dv}{dx}\right\}$

The above proves $\left(\frac{u}{v}\right)\prime =\frac{u\prime v-uv\prime }{{v}^{2}}$

Intuitive understanding of $\left(\frac{u}{v}\right)\prime =\frac{u\prime v-uv\prime }{{v}^{2}}$

This can be rewritten as
$\left(\frac{u}{v}\right)\prime =$$\left(u\prime ×\left(\frac{1}{v}\right)$$+u×\left(-\frac{1}{{v}^{2}}\right)×v\prime$

The first part of the result is :
fixing $\frac{1}{v}$$\frac{1}{v}$ as a constant, rate of change of $u$$u$ is taken

The second part of the result is :
fixing $u$$u$ as a constant, rate of change of $\frac{1}{v}$$\frac{1}{v}$ is taken.
Rate of change of $\frac{1}{v}$$\frac{1}{v}$ is negative of rate of change of $v$$v$ and divided by ${v}^{2}$${v}^{2}$.

Given $y=\frac{\mathrm{sin}x}{{x}^{2}}$$y = \frac{\sin x}{{x}^{2}}$ what is $\frac{dy}{dx}$$\frac{\mathrm{dy}}{\mathrm{dx}}$? Note: $\frac{d}{dx}{x}^{2}=2x$$\frac{d}{\mathrm{dx}} {x}^{2} = 2 x$
$\frac{d}{dx}\mathrm{sin}x=\mathrm{cos}x$$\frac{d}{\mathrm{dx}} \sin x = \cos x$

The answer is "$\frac{{x}^{2}\mathrm{cos}x-2x\mathrm{sin}x}{{x}^{4}}$$\frac{{x}^{2} \cos x - 2 x \sin x}{{x}^{4}}$"

$\frac{dy}{dx}$$\frac{\mathrm{dy}}{\mathrm{dx}}$
$=\frac{d}{dx}\left[\frac{\mathrm{sin}x}{{x}^{2}}\right]$$= \frac{d}{\mathrm{dx}} \left[\frac{\sin x}{{x}^{2}}\right]$
applying quotient law of derivative $=\frac{{x}^{2}\frac{d}{dx}\mathrm{sin}x-\mathrm{sin}x\frac{d}{dx}{x}^{2}}{{x}^{4}}$$= \frac{{x}^{2} \frac{d}{\mathrm{dx}} \sin x - \sin x \frac{d}{\mathrm{dx}} {x}^{2}}{{x}^{4}}$
$=\frac{{x}^{2}\mathrm{cos}x-2x\mathrm{sin}x}{{x}^{4}}$$= \frac{{x}^{2} \cos x - 2 x \sin x}{{x}^{4}}$

summary

Derivative of Quotient: $\left(\frac{u}{v}\right)\prime =\frac{u\prime v-uv\prime }{{v}^{2}}$

composite

Finding derivative of composite function.
Given $v\left(u\right)$$v \left(u\right)$ and $u\left(x\right)$$u \left(x\right)$ find $\frac{dv}{dx}$$\frac{\mathrm{dv}}{\mathrm{dx}}$

$\frac{dv}{dx}$$\frac{\mathrm{dv}}{\mathrm{dx}}$

$=\underset{\Delta x\to 0}{lim}\frac{v\left(u\left(x+\Delta x\right)\right)-v\left(u\left(x\right)\right)}{\Delta x}$$= {\lim}_{\Delta x \to 0} \frac{v \left(u \left(x + \Delta x\right)\right) - v \left(u \left(x\right)\right)}{\Delta x}$

multiplying and dividing by $u\left(x+\Delta x\right)-u\left(x\right)$$u \left(x + \Delta x\right) - u \left(x\right)$

$=\underset{\Delta x\to 0}{lim}\left[$$\frac{v\left(u\left(x+\Delta x\right)\right)-v\left(u\left(x\right)\right)}{u\left(x+\Delta x\right)-u\left(x\right)}$$\frac{v \left(u \left(x + \Delta x\right)\right) - v \left(u \left(x\right)\right)}{u \left(x + \Delta x\right) - u \left(x\right)}$$×\frac{u\left(x+\Delta x\right)-u\left(x\right)}{\Delta x}$$\times \frac{u \left(x + \Delta x\right) - u \left(x\right)}{\Delta x}$$\right]$

Substituting $u\left(x+\Delta x\right)-u\left(x\right)=\Delta u$$u \left(x + \Delta x\right) - u \left(x\right) = \Delta u$
If $\Delta x\to 0$$\Delta x \to 0$, then $\Delta u\to 0$$\Delta u \to 0$
with continuity and differentiability conditions on $v$$v$ and $u$$u$.

$=\underset{\Delta u\to 0}{lim}\frac{v\left(u\left(x+\Delta x\right)\right)-v\left(u\left(x\right)\right)}{\Delta u}$$= {\lim}_{\Delta u \to 0} \frac{v \left(u \left(x + \Delta x\right)\right) - v \left(u \left(x\right)\right)}{\Delta u}$$\underset{\Delta x\to 0}{lim}\frac{u\left(x+\Delta x\right)-u\left(x\right)}{\Delta x}$${\lim}_{\Delta x \to 0} \frac{u \left(x + \Delta x\right) - u \left(x\right)}{\Delta x}$

$=\frac{dv}{du}\frac{du}{dx}$$= \frac{\mathrm{dv}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$

The above proves $\frac{dv}{dx}=\frac{dv}{du}\frac{du}{dx}$$\frac{\mathrm{dv}}{\mathrm{dx}} = \frac{\mathrm{dv}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$

Derivative of composite function can be extended to multi-level functions.

$\frac{df}{dx}=\frac{df}{dg}\frac{dg}{dh}\frac{dh}{dv}\frac{dv}{dx}$$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{dg}} \frac{\mathrm{dg}}{\mathrm{dh}} \frac{\mathrm{dh}}{\mathrm{dv}} \frac{\mathrm{dv}}{\mathrm{dx}}$

This is also called chain rule of differentiation.

Intuitive understanding of chain rule :


In $v\left(u\left(x\right)\right)$$v \left(u \left(x\right)\right)$, $v\left(u\right)$$v \left(u\right)$ is called outer-function and $u\left(x\right)$$u \left(x\right)$ is called inner-function. In a composite function,

the change in variable causes a change in inner-function and that change is the rate of change of inner-function.

the change in inner-function, in turn, causes change in outer-function. This change, rate of change of outer-function, is with respect to the inner-function.

Thus rate of change of outer function with respect to variable $=$$=$ rate of change of inner-function to the variable $×$$\times$ rate of change of outer-function to the inner-function.

Given $y=\mathrm{sin}\left({x}^{2}\right)$$y = \sin \left({x}^{2}\right)$, what is $\frac{dy}{dx}$$\frac{\mathrm{dy}}{\mathrm{dx}}$?
Note: $\frac{d}{dx}{x}^{2}=2x$$\frac{d}{\mathrm{dx}} {x}^{2} = 2 x$
$\frac{d}{dx}\mathrm{sin}x=\mathrm{cos}x$$\frac{d}{\mathrm{dx}} \sin x = \cos x$

The answer is "$2x\mathrm{cos}\left({x}^{2}\right)$$2 x \cos \left({x}^{2}\right)$"

$\frac{dy}{dx}$$\frac{\mathrm{dy}}{\mathrm{dx}}$

taking $u={x}^{2}$$u = {x}^{2}$
$=\frac{d}{du}\mathrm{sin}u×\frac{d}{dx}{x}^{2}$$= \frac{d}{\mathrm{du}} \sin u \times \frac{d}{\mathrm{dx}} {x}^{2}$

applying law of derivative
$=\mathrm{cos}u×2x$$= \cos u \times 2 x$
$=2x\mathrm{cos}\left({x}^{2}\right)$$= 2 x \cos \left({x}^{2}\right)$

Derivative of Composite Function : $\frac{dv}{dx}=\frac{dv}{du}\frac{du}{dx}$$\frac{\mathrm{dv}}{\mathrm{dx}} = \frac{\mathrm{dv}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$

parametric form

Finding derivative of function given in parametric form.
$y=f\left(r\right)$$y = f \left(r\right)$ and $x=g\left(r\right)$$x = g \left(r\right)$
$\frac{dy}{dx}=?$

$\frac{dy}{dx}$$\frac{\mathrm{dy}}{\mathrm{dx}}$ $=\underset{\Delta x\to 0}{lim}\left[\Delta y÷\Delta x\right]$$= {\lim}_{\Delta x \to 0} \left[\Delta y \div \Delta x\right]$

Note that change in $x$$x$ will reflect as change in $r$$r$.
$\Delta x=g\left(r+\Delta r\right)-g\left(r\right)$$\Delta x = g \left(r + \Delta r\right) - g \left(r\right)$
$\Delta y=f\left(r+\Delta r\right)-f\left(r\right)$$\Delta y = f \left(r + \Delta r\right) - f \left(r\right)$

$\frac{df}{dg}$$\frac{\mathrm{df}}{\mathrm{dg}}$
$=\underset{\Delta r\to 0}{lim}\left[$$\left(f\left(r+\Delta r\right)-f\left(r\right)\right)$$\left(f \left(r + \Delta r\right) - f \left(r\right)\right)$$÷\left(g\left(r+\Delta r\right)-g\left(r\right)\right)\right]$

dividing numerator and denominator by $\Delta r$$\Delta r$

$=\underset{\Delta r\to 0}{lim}\left[$$\left(f\left(r+\Delta r\right)-f\left(r\right)\right)/\Delta r$$\left(f \left(r + \Delta r\right) - f \left(r\right)\right) / \Delta r$$÷\left(g\left(r+\Delta r\right)-g\left(r\right)\right)/\Delta r$$\div \left(g \left(r + \Delta r\right) - g \left(r\right)\right) / \Delta r$$\right]$

with continuity and differentiability conditions on $u$$u$ and $a$$a$
$=\underset{\Delta r\to 0}{lim}\left[$$\left(f\left(r+\Delta r\right)-f\left(r\right)\right)/\Delta r\right]$$÷\underset{\Delta r\to 0}{lim}\left[\left(g\left(r+\Delta r\right)-g\left(r\right)\right)/\Delta r$$\right]$

$=\frac{df/dr}{dg/dr}$$= \frac{\mathrm{df} / \mathrm{dr}}{\mathrm{dg} / \mathrm{dr}}$

$=\frac{dy/dr}{dx/dr}$$= \frac{\mathrm{dy} / \mathrm{dr}}{\mathrm{dx} / \mathrm{dr}}$

The above proves $\frac{dv}{du}=dv/dx÷du/dx$$\frac{\mathrm{dv}}{\mathrm{du}} = \mathrm{dv} / \mathrm{dx} \div \mathrm{du} / \mathrm{dx}$

Given $y=\mathrm{sin}r$$y = \sin r$ and $x={r}^{2}$$x = {r}^{2}$, what is $\frac{dy}{dx}$$\frac{\mathrm{dy}}{\mathrm{dx}}$?
Note: $\frac{d}{dp}{p}^{2}=2p$$\frac{d}{\mathrm{dp}} {p}^{2} = 2 p$
$\frac{d}{dp}\mathrm{sin}p=\mathrm{cos}p$$\frac{d}{\mathrm{dp}} \sin p = \cos p$?

The answer is "$\frac{\mathrm{cos}r}{2r}$$\frac{\cos r}{2 r}$"

$\frac{dy}{dx}$$\frac{\mathrm{dy}}{\mathrm{dx}}$
$\frac{dy/dr}{dx/dr}$$\frac{\mathrm{dy} / \mathrm{dr}}{\mathrm{dx} / \mathrm{dr}}$
$=\frac{\mathrm{cos}r}{2r}$$= \frac{\cos r}{2 r}$
as $\frac{dy}{dr}=\mathrm{cos}r$$\frac{\mathrm{dy}}{\mathrm{dr}} = \cos r$ and $\frac{dx}{dr}=2r$$\frac{\mathrm{dx}}{\mathrm{dr}} = 2 r$

Intuitive understanding of $\frac{dv}{du}=\frac{dv/dx}{du/dx}$$\frac{\mathrm{dv}}{\mathrm{du}} = \frac{\mathrm{dv} / \mathrm{dx}}{\mathrm{du} / \mathrm{dx}}$ Rate of change of a function $v$$v$ with respect to another function $u$$u$ can be computed by the change in the common variable $x$$x$. The change in the common variable $x$$x$ causes change in $v$$v$ as well as in $u$$u$. This leads to the given formula.

summary

Derivative of function in parametric form: $\frac{dv}{du}=\frac{dv/dx}{du/dx}$$\frac{\mathrm{dv}}{\mathrm{du}} = \frac{\mathrm{dv} / \mathrm{dx}}{\mathrm{du} / \mathrm{dx}}$

summary

Derivative of a Multiple:
$\left(au\right)\prime =au\prime$
Derivative of a multiple of a function is multiple of the derivative of the function.

Derivative of Sum or Difference:
$\left(u+v\right)\prime =u\prime +v\prime$

$\left(u-v\right)\prime =u\prime -v\prime$
Derivative of a sum or difference is the sum or difference of derivatives.

Derivative of Product: $\left(uv\right)\prime =u\prime v+uv\prime$
Derivative of product of two function is sum of derivatives of the functions scaled by the value of other function

Derivative of Quotient: $\left(\frac{u}{v}\right)\prime =\frac{u\prime v-uv\prime }{{v}^{2}}$

Derivative of function in parametric form: $\frac{dv}{du}=\frac{dv/dx}{du/dx}$$\frac{\mathrm{dv}}{\mathrm{du}} = \frac{\mathrm{dv} / \mathrm{dx}}{\mathrm{du} / \mathrm{dx}}$

Outline

The outline of material to learn "Differential Calculus" is as follows.

•   Detailed outline of Differential Calculus

→   Application Scenario

→   Differentiation in First Principles

→   Graphical Meaning of Differentiation

→   Differntiability

→   Algebra of Derivatives

→   Standard Results