Differential Calculus : Derivatives of Algebraic Expressions

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Derivatives of Algebraic Expressions

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Overview

In this page, Derivatives of the following are explained.

→ standard functions in algebraic expressions

→ trigonometric functions sin, cos, tan etc.

→ inverse trigonometric functions such as $\arcsin$ $arcsin$ , $\arccos$ $arccos$ , etc.

→ exponents and logarithmic functions such as $e^{x}$ $e^x$ , $a^{x}$ $a^x$ , and $\ln x$ $ln x$

constant

Finding the derivative of $y = a$ $y=a$ in first principles:

$\frac{d y}{d x}$ $(dy)/(dx)$

$= lim_{δ \to 0} [f (x + δ)$ $=lim_(delta->0) [color(coral)(f(x+delta))$ $- f (x)] / δ$ $- color(deepskyblue)(f(x))]//delta$

$= lim_{δ \to 0} [a$ $=lim_(delta->0) [color(coral)(a)$ $- a] / δ$ $- color(deepskyblue)(a)]//delta$

$= lim_{δ \to 0} [0] / δ$ $=lim_(delta->0) [0]//delta$

The above proves
$\frac{d}{d x} (a) = 0$ $d/(dx) (a) = 0$
Note that $δ$ $delta$ is very small value and the numerator is $0$ $0$ . So the limit evaluates to $0$ $0$ .

exponent

Finding the derivative of $y = x^{n}$ $y=x^n$ in first principles:

$\frac{d y}{d x}$ $(dy)/(dx)$

$= lim_{δ \to 0} [f (x + δ)$ $=lim_(delta->0) [color(coral)(f(x+delta))$ $- f (x)] / δ$ $- color(deepskyblue)(f(x))]//delta$

$= lim_{δ \to 0} [{(x + δ)}^{n}$ $=lim_(delta->0) [color(coral)((x+delta)^n)$ $- x^{n}] / δ$ $- color(deepskyblue)(x^n)]//delta$

using binomial expression for ${(x + δ)}^{n}$ $(x+delta)^n$ $= lim_{δ \to 0} [(x^{n} + n δ x^{n - 1}$ $=lim_(delta->0) [color(coral)((x^n +ndelta x^(n-1)$ $+^{n} C_{2} δ^{2} x^{n - 2}$ $+ ^nC_2 delta^2x^(n-2)$ $+ \dots + δ^{n})$ $+ cdots+delta^n)$ $- x^{n}] / δ$ $- color(deepskyblue)(x^n)]//delta$

canceling $x^{n}$ $color(coral)(x^n)$ and $- x^{n}$ $color(deepskyblue)(-x^n)$
$= lim_{δ \to 0} [(n δ x^{n - 1}$ $=lim_(delta->0) [color(coral)((ndelta x^(n-1)$ $+^{n} C_{2} δ^{2} x^{n - 2}$ $+ ^nC_2 delta^2x^(n-2)$ $+ \dots + δ^{n})$ $+ cdots+delta^n)$ $] / δ$ $]//delta$

dividing all terms by $δ$ $delta$
$= lim_{δ \to 0} [(n x^{n - 1}$ $=lim_(delta->0) [color(coral)((nx^(n-1)$ $+^{n} C_{2} δ^{1} x^{n - 2}$ $+ ^nC_2 delta^1x^(n-2)$ $+ \dots + δ^{n - 1})$ $+ cdots+delta^(n-1))$ $]$ $]$

applying limit $δ \to 0$ $delta->0$
The $δ$ $delta$ is substituted as $0$ $0$ and only one term is non zero. The above proves
$\frac{d}{d x} (x^{n}) = n x^{n - 1}$ $d/(dx) (x^n) = nx^(n-1)$ ".

summary

Derivatives of Algebraic Expressions :
Derivative of a constant
$\frac{d}{d x} a = 0$ $d/(dx) a = 0$
rate of change of constant is 0

Derivative of power:
$\frac{d}{d x} x^{n} = n x^{n - 1}$ $d/(dx) x^n = n x^(n-1)$
small change results in binomial expansion and only one factor remains

example

Find the derivative of $x^{3} + 4 x^{2} - 3 x - 2$ $x^3+4x^2-3x-2$
The answer is " $3 x^{2} + 8 x - 3$ $3x^2+8x-3$ "

sine

Finding the derivative of $y = \sin x$ $y=sin x$ in first principles:

$\frac{d y}{d x}$ $(dy)/(dx)$

$= lim_{δ \to 0} [f (x + δ)$ $=lim_(delta->0) [color(coral)(f(x+delta))$ $- f (x)] / δ$ $- color(deepskyblue)(f(x))]//delta$

$= lim_{δ \to 0} [\sin (x + δ)$ $=lim_(delta->0) [color(coral)(sin(x+delta)$ $- \sin x] / δ$ $- color(deepskyblue)(sinx)]//delta$

$= lim_{δ \to 0} [\sin x \cos δ + \cos x \sin δ$ $=lim_(delta->0) [color(coral)(sinx cos delta + cos x sin delta)$ $- \sin x] / δ$ $-color(deepskyblue)(sinx)]//delta$

$= lim_{δ \to 0} \frac{\cos x \sin δ}{δ}$ $= lim_(delta->0) [color(coral)(cos x sin delta)]/delta$ $- lim_{δ \to 0} \frac{\sin x - \sin x \cos δ}{δ}$ $- lim_(delta->0) (color(deepskyblue)(sin x) - color(coral)(sin x cos delta))/delta$

$= \cos x lim_{δ \to 0} \frac{\sin δ}{δ}$ $= cos x lim_(delta->0) (sin delta)/delta$ $- \sin x lim_{δ \to 0} \frac{1 - \cos δ}{δ}$ $- sin x lim_(delta->0)(1 - cos delta)/delta$

applying the standard limits
$= \cos x \times 1 - \sin x \times 0$ $= cos x xx 1 - sin x xx 0$

The above proves
$\frac{d}{d x} \sin x = \cos x$ $d/(dx) sin x = cos x$

cosine

Finding the derivative of $y = \cos x$ $y=cos x$ :

$\frac{d}{d x} \cos x$ $(d)/(dx) cos x$

$= \frac{d}{d x} \sin (\frac{π}{2} - x)$ $=d/(dx) sin(pi/2-x)$

applying chain rule of differentiation with $u = \frac{π}{2} - x$ $u=pi/2-x$
$= \frac{d}{d u} \sin (u) \frac{d}{d x} (\frac{π}{2} - x)$ $=d/(du)sin(u) d/(dx) (pi/2-x)$

$= \cos (u) \times (- 1)$ $=cos(u) xx (-1)$

$= - \cos (\frac{π}{2} - x)$ $=-cos(pi/2-x)$

$= - \sin (x)$ $=-sin(x)$

The above proves
$\frac{d}{d x} \cos x = - \sin x$ $d/(dx) cos x = -sin x$

tangent

Finding the derivative of $y = \tan x$ $y=tan x$

$(\tan x)'$ $(tan x)′$

$= (\frac{\sin x}{\cos x})'$ $=((sin x)/(cos x))′$

applying quotient rule of differentiation
$= \frac{\cos x (\sin x)' - \sin x (\cos x)'}{\cos^{2} x}$ $=(cos x(sin x)′ - sin x(cos x)′)/(cos^2 x)$

$= \frac{\cos x (\cos x) - \sin x (- \sin x)}{\cos^{2} x}$ $=(cos x(cos x) - sin x(-sin x))/(cos^2 x)$

$= \frac{\cos^{2} x + \sin^{2} x}{\cos^{2} x}$ $=(cos^2 x + sin^2 x)/(cos^2 x)$

$= \frac{1}{\cos^{2} x}$ $=1/(cos^2 x)$

$= \sec^{2} (x)$ $=sec^2(x)$

The above proves
$\frac{d}{d x} \tan x = \sec^{2} x$ $d/(dx) tan x = sec^2 x$

cotangent

Finding the derivative of $y = \cot x$ $y=cot x$

$(\cot x)'$ $(cot x)′$

$= (\frac{\cos x}{\sin x})'$ $=((cos x)/(sin x))′$

applying quotient rule of differentiation
$= \frac{\sin x (\cos x)' - \cos x (\sin x)'}{\sin^{2} x}$ $=(sin x(cos x)′ - cos x(sin x)′)/(sin^2 x)$

$= \frac{\sin x (- \sin x) - \cos x (\cos x)}{\sin^{2} x}$ $=(sin x(-sin x) - cos x(cos x))/(sin^2 x)$

$= \frac{- \sin^{2} x - \cos^{2} x}{\sin^{2} x}$ $=(-sin^2 x - cos^2 x)/(sin^2 x)$

$= \frac{- 1}{\sin^{2} x}$ $=(-1)/(sin^2 x)$

$= - \csc^{2} (x)$ $=-csc^2(x)$

The above proves
$\frac{d}{d x} \cot x = - \csc^{2} x$ $d/(dx) cot x = -csc^2 x$

secant

Finding the derivative of $y = \sec x$ $y=sec x$

$(\sec x)'$ $(sec x)′$

$= (\frac{1}{\cos x})'$ $=(1/(cos x))′$

applying quotient rule of differentiation
$= \frac{\cos x (1)' - (\cos x)'}{\cos^{2} x}$ $=(cos x(1)′ - (cos x)′)/(cos^2 x)$

$= \frac{\cos x (0) - (- \sin x)}{\cos^{2} x}$ $=(cos x(0) - (-sin x))/(cos^2 x)$

$= \frac{\sin x}{\cos^{2} x}$ $=(sin x)/(cos^2 x)$

$= \sec x \tan x$ $=sec x tan x$

The above proves
$\frac{d}{d x} \sec x = \sec x \tan x$ $d/(dx) sec x = sec x tan x$

cosecant

Finding the derivative of $y = \csc x$ $y=csc x$

$(\csc x)'$ $(csc x)′$

$= (\frac{1}{\sin x})'$ $=(1/(sin x))′$

applying quotient rule of differentiation
$= \frac{\sin x (1)' - (\sin x)'}{\sin^{2} x}$ $=(sin x(1)′ - (sin x)′)/(sin^2 x)$

$= \frac{\sin x (0) - (\cos x)}{\sin^{2} x}$ $=(sin x(0) - (cos x))/(sin^2 x)$

$= \frac{- \cos x}{\sin^{2} x}$ $=(-cos x)/(sin^2 x)$

$= - \csc x \cot x$ $=-csc x cot x$

The above proves
$\frac{d}{d x} \csc x = - \csc x \cot x$ $d/(dx) csc x =- csc x cot x$

summary

Derivatives of Trigonometric Functions:
$\frac{d}{d x} \sin x = \cos x$ $d/(dx) sin x = cos x$

$\frac{d}{d x} \cos x = - \sin x$ $d/(dx) cos x = -sin x$
$\cos x = \sin (\frac{π}{2} - x)$ $cos x = sin(pi/2 -x)$ and the negative sign of $x$ $x$ is carried in the result

$\frac{d}{d x} \tan x = \sec^{2} x$ $d/(dx) tan x = sec^2 x$
$\tan = \frac{\sin}{\cos}$ $tan = (sin)/(cos)$ and results in $\frac{1}{\cos^{2}}$ $1/(cos^2)$

$\frac{d}{d x} \cot x = - \csc^{2} x$ $d/(dx) cot x = -csc^2 x$
$\cot = \tan (\frac{π}{2} - x)$ $cot = tan(pi/2-x)$ and the negative sign of x is carried

$\frac{d}{d x} \sec x = \sec x \tan x$ $d/(dx) sec x = sec x tan x$
$\sec = \frac{1}{\cos}$ $sec = 1/cos$ and results in $\frac{\sin}{\cos^{2}}$ $sin/(cos^2)$

$\frac{d}{d x} \csc x = - \csc x \cot x$ $d/(dx) csc x = -csc x cot x$
$\csc = \sec (\frac{π}{2} - x)$ $csc = sec(pi/2-x)$ and the negative sign of x is carried

example

Find the derivative of $\tan x - x$ $tan x -x$
The answer is " $\tan^{2} x$ $tan^2 x$ ".

$\tan x - x$ $tan x -x$
$\sec^{2} x - 1$ $sec^2 x -1$
$\tan^{2} x$ $tan^2x$

inverse sine

Finding the derivative of $y = \arcsin x$ $y = arcsinx$ :
$y = \arcsin x$ $y= arcsin x$
$\sin y = x$ $sin y = x$
differentiate this
$\cos y \frac{d y}{d x} = 1$ $cos y (dy)/(dx) = 1$

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

The above proves
$\frac{d}{d x} \arcsin x = \frac{1}{\sqrt{1 - x^{2}}}$ $d/(dx)arcsinx = 1/(sqrt(1-x^2))$

inverse cosine

Finding the derivative of $y = \arccos x$ $y = arccosx$ :
$y = \arccos x$ $y=arccosx$
$\cos y = x$ $cos y = x$
differentiate this
$- \sin y \frac{d y}{d x} = 1$ $-sin y (dy)/(dx) = 1$

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

The above proves
$\frac{d}{d x} \arccos x = \frac{- 1}{\sqrt{1 - x^{2}}}$ $d/(dx)arccosx = (-1)/(sqrt(1-x^2))$

inverse tangent

Finding the derivative of $y = \arctan x$ $y = arctanx$ :
$y = \arctan x$ $y=arctanx$
$\tan y = x$ $tan y = x$
differentiate this
$\sec^{2} y \frac{d y}{d x} = 1$ $sec^2y (dy)/(dx) = 1$

$\frac{d y}{d x} = \frac{1}{\sec^{2} y}$ $(dy)/(dx) = 1/(sec^2y)$
$= \frac{1}{\sqrt{1 + \tan^{2} y}}$ $= (1)/sqrt(1+tan^2 y)$
$= \frac{1}{\sqrt{1 + x^{2}}}$ $= (1)/sqrt(1+x^2)$

The above proves
$\frac{d}{d x} \arctan x = \frac{1}{1 + x^{2}}$ $d/(dx)arctanx = 1/(1+x^2)$

inverse secant

Finding the derivative of $y = a r c \sec x$ $y = arc secx$ :
$y = a r c \sec x$ $y=arc secx$
$y$ $y$ is in first or second quadrant.
$\sec y = x$ $sec y = x$
differentiate this
$\sec y \tan y \frac{d y}{d x} = 1$ $sec y tan y (dy)/(dx) = 1$
$\frac{d y}{d x} = \frac{1}{\sec y \tan y}$ $(dy)/(dx) = 1/(sec y tan y)$
$\sec y \tan y$ $sec y tan y$ is always positive for values in first and second quadrant.
$\frac{d y}{d x} = \frac{1}{\sec y \sqrt{\sec^{2} y - 1}}$ $(dy)/(dx) = 1/(sec y sqrt(sec^2 y-1))$
$= \frac{1}{| x | \sqrt{x^{2} - 1}}$ $= 1/(|x|sqrt(x^2-1))$

The above proves
$\frac{d}{d x} a r c \sec x = \frac{1}{| x | \sqrt{x^{2} - 1}}$ $d/(dx)arc secx = 1/(|x|sqrt(x^2-1))$

inverse cosecant

Finding the derivative of $y = a r c \csc x$ $y = arc cscx$ :
$y = a r c \csc x$ $y=arc cscx$
$y$ $y$ is in range $- \frac{π}{2} \leq y \leq \frac{π}{2}$ $-pi/2<=y<=pi/2$ .
$\csc y = x$ $csc y = x$
differentiate this
$- \csc y \cot y \frac{d y}{d x} = 1$ $-csc y cot y (dy)/(dx) = 1$

$\frac{d y}{d x} = \frac{- 1}{\csc y \cot y}$ $(dy)/(dx) = (-1)/(csc y cot y)$
$\csc y \cot y$ $csc y cot y$ is always positive for values in range $- \frac{π}{2} \leq y \leq \frac{π}{2}$ $-pi/2<=y<=pi/2$
$= \frac{- 1}{\csc y \sqrt{\csc^{2} y - 1}}$ $= (-1)/(csc y sqrt(csc^2 y-1))$
$= \frac{- 1}{| x | \sqrt{x^{2} - 1}}$ $= (-1)/(|x|sqrt(x^2-1))$

The above proves
$\frac{d}{d x} a r c \csc x = \frac{- 1}{| x | \sqrt{x^{2} - 1}}$ $d/(dx)arc cscx = (-1)/(|x|sqrt(x^2-1))$

inverse cotangent

Finding the derivative of $y = a r c \cot x$ $y = arc cotx$ :
$y = a r c \cot x$ $y=arc cotx$
$\cot y = x$ $cot y = x$
differentiate this
$- \csc^{2} y \frac{d y}{d x} = 1$ $-csc^2y (dy)/(dx) = 1$

$\frac{d y}{d x} = \frac{- 1}{\csc^{2} y}$ $(dy)/(dx) = (-1)/(csc^2y)$
$= \frac{- 1}{\sqrt{1 + \tan^{2} y}}$ $= (-1)/sqrt(1+tan^2 y)$
$= \frac{- 1}{\sqrt{1 + x^{2}}}$ $= (-1)/sqrt(1+x^2)$

The above proves
$\frac{d}{d x} a r c \cot x = \frac{- 1}{1 + x^{2}}$ $d/(dx) arc cotx = (-1)/(1+x^2)$

summary

Derivatives of inverse trigonometric functions : $trig fn (y) = x$ $text(trig fn) (y)=x$ , so derivative in terms of x

$\frac{d}{d x} \arcsin x = \frac{1}{\sqrt{1 - x^{2}}}$ $d/(dx) arcsinx = 1/(sqrt(1-x^2))$

$\frac{d}{d x} \arccos x = \frac{- 1}{\sqrt{1 - x^{2}}}$ $d/(dx) arccosx = (-1)/(sqrt(1-x^2))$

$\frac{d}{d x} \arctan x = \frac{1}{1 + x^{2}}$ $d/(dx) arctanx = 1/(1+x^2)$

$\frac{d}{d x} a r c \sec x = \frac{1}{| x | \sqrt{x^{2} - 1}}$ $d/(dx) arc secx = 1/(|x|sqrt(x^2-1))$

$\frac{d}{d x} a r c \csc x = \frac{- 1}{| x | \sqrt{x^{2} - 1}}$ $d/(dx) arc cscx = (-1)/(|x|sqrt(x^2-1))$

$\frac{d}{d x} a r c \cot x = \frac{- 1}{1 + x^{2}}$ $d/(dx) arc cotx = (-1)/(1+x^2)$

example

Find the derivative of $(1 + x^{2}) \arctan x$ $(1+x^2)arctan x$
The answer is " $1 + 2 x \arctan x$ $1+2x arctan x$ "

$[(1 + x^{2}) \arctan x]'$ $[ (1+x^2)arctan x]′$
$(1 + x^{2}) (\arctan x)' + (1 + x^{2})' \arctan x$ $(1+x^2) (arctan x)′ + (1+x^2)′ arctan x$
$(1 + x^{2}) (\frac{1}{1 + x^{2}}) + 2 x \arctan x$ $(1+x^2) (1/(1+x^2)) + 2x arctan x$
$1 + 2 x \arctan x$ $1+2x arctan x$

exponent of e

Finding the derivative of $y = e^{x}$ $y=e^(x)$ in first principles:

$\frac{d}{d x} e^{x}$ $d/(dx) e^(x)$

$= lim_{δ \to 0} [e^{x + δ}$ $=lim_(delta->0) [color(coral)(e^(x+delta)$ $- e^{x}] / δ$ $- color(deepskyblue)(e^x)]//delta$

$= lim_{δ \to 0} [e^{x} \times e^{δ}$ $=lim_(delta->0) [color(coral)(e^x xx e^(delta))$ $- e^{x}] / δ$ $- color(deepskyblue)(e^x)]//delta$

$= lim_{δ \to 0} e^{x} [e^{δ}$ $=lim_(delta->0) e^x[color(coral)( e^(delta))$ $- 1] / δ$ $- color(deepskyblue)(1)]//delta$

applying the standard limit $lim_{p \to 0} (e^{p} - 1) / p = 1$ $lim_(p->0)(e^p - 1)//p = 1$

$= e^{x}$ $= e^(x)$

The above proves
$\frac{d}{d x} e^{x} = e^{x}$ $d/(dx) e^(x) = e^(x)$

logarithm base e

Finding the derivative of $y = \ln x$ $y=ln x$ :

$y = \ln x$ $y=ln x$

$e^{y} = x$ $e^y=x$

differentiating the equation
$\frac{d}{d x} e^{y} = 1$ $(d)/(dx)e^y=1$

applying chain rule $\frac{d}{d y} e^{y} \frac{d y}{d x} = 1$ $(d)/(dy)e^y (dy)/(dx)=1$

$e^{y} \frac{d y}{d x} = 1$ $e^y (dy)/(dx)=1$

$x \frac{d y}{d x} = 1$ $x (dy)/(dx)=1$

$\frac{d y}{d x} = \frac{1}{x}$ $(dy)/(dx)=1/x$

The above proves
$\frac{d}{d x} \ln x = \frac{1}{x}$ $d/(dx) ln x = 1/x$

exponent

Finding the derivative of $y = a^{x}$ $y=a^x$ :

substituting $a = e^{\ln a}$ $a=e^(ln a)$

$y = e^{x \ln a}$ $y=e^(x ln a )$

differentiating the equation
$\frac{d y}{d x} = \frac{d}{d x} e^{x \ln a}$ $(dy)/(dx)=(d)/(dx) e^(x ln a )$

applying chain rule with $u = x \ln a$ $u=x ln a$
$\frac{d y}{d x} = \frac{d}{d u} e^{u} \frac{d}{d x} (x \ln a)$ $(dy)/(dx)=(d)/(du) e^u (d)/(dx)(x ln a )$

$\frac{d y}{d x} = e^{u} \times \ln a$ $(dy)/(dx)= e^u xx ln a$

$\frac{d y}{d x} = e^{x \ln a} \times \ln a$ $(dy)/(dx)= e^(x ln a ) xx ln a$

substituting $e^{\ln a} = a$ $e^(ln a)=a$

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

The above proves
$\frac{d}{d x} a^{x} = a^{x} \ln a$ $d/(dx) a^x = a^x ln a$

summary

$\frac{d}{d x} e^{x} = e^{x}$ $d/(dx) e^x = e^x$
definition of $e$ $e$ is rate of change is proportional to itself

$\frac{d}{d x} a^{x} = a^{x} \ln a$ $d/(dx) a^x = a^x ln a$
$a$ $a$ equals $e^{\ln a}$ $e^(lna)$

$\frac{d}{d x} \ln x = \frac{1}{x}$ $d/(dx) ln x = 1/x$
natural log is inverse of $e$ $e$ power

example

What is the derivative of $\log_{10} x$ $log_(10) x$ ?
Note: Use the identity $\log_{10} x = \frac{\log_{e} x}{\log_{e} 10}$ $log_(10) x = (log_e x)/(log_e 10)$

The answer is " $\frac{1}{x \ln 10}$ $1/(xln 10)$ ".
$\frac{d}{d x} \log_{10} x$ $d/(dx)log_10 x$
$= \frac{d}{d x} \frac{\log_{e} x}{\log_{e} 10}$ $= d/(dx)(log_e x)/(log_e 10)$
$= \frac{1}{\log_{e} 10} \frac{d}{d x} \log_{e} x$ $= 1/(log_e 10) d/(dx) log_e x$
$= \frac{1}{\log_{e} 10} \times \frac{1}{x}$ $=1/(log_e 10) xx 1/x$
$= \frac{1}{x \log_{e} 10}$ $=1/(xlog_e 10)$

What is the derivative of $e^{a x}$ $e^(ax)$ ?

The answer is " $a e^{a x}$ $ae^(ax)$ ". Applying chain rule with $u = a x$ $u=ax$ ,
$\frac{d}{d u} e^{u} \frac{d}{d x} a x$ $d/(du) e^u d/(dx) ax$
$e^{u} \times a$ $e^u xx a$
$a e^{a x}$ $ae^(ax)$

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