derivative , differentitation

guide to differentiation

A derivative is an operation that quantifies the sensitivity of change of a function’s output as its input changes

The act of finding a derivative is called differentiation.
Differentiation is the core of differential calculus.

syntax

there are a few ways derivatives are presented.

if we have an expression $x^3$, its derivative is shown as $\frac{d}{dx}[x^3]$
if we have a function $f(x)=x^3$, the derivative of the function is $f^{\prime }(x)=\frac{d}{dx}[x^3]$
if we have an equation $y=x^3$, the derivative of the equation is $\frac{dy}{dx}=\frac{d}{dx}[x^3]$
notice how differentiation is an operation so it can be done to both sides of the equation

the derivative of a function $f(x)$ are point $x=c$ is the slope of tangent line at that point

finding the derivative is called differentiation. for a function $f(x)$ we can find its derivative with:

$f^{\prime }(x)=\underset{x\to c}{\lim }\frac{f(x)-f(c)}{x-c}$

this definition is explained in proof of finding derivative

differentiability

in calculus, we have a way to tell if you can perform a derivative on a function

for a point $x=c$ on function $f$: $f$ is not differentiable if $f$ is not continuous at $x=c$, or if $f$ has a sharp turn at $x=c$. in order for $f$ to be continuous at $x=c$, $\underset{x\to c}{\lim }\frac{f(x)-f(c)}{x-c}$ must exist.

if differentiable, then it must be continuous.

derivative rules

constant rule

$\frac{d}{dx}\left[c\right]=0$

variable rule

$\frac{d}{dx}\left[x\right]=1$

power rule

$\frac{d}{dx}[x^n]=n\cdot x^{n-1}$

constant multiple rule

$\frac{d}{dx}\left[c\cdot f(x)\right]=c\cdot \frac{d}{dx}\left[f(x)\right]$

sum rule

$\frac{d}{dx}\left[f(x)+g(x)\right]=\frac{d}{dx}\left[f(x)\right]+\frac{d}{dx}\left[g(x)\right]$

product rule

$\frac{d}{dx}\left[f(x)g(x)\right]=f^{\prime }(x)g(x)+f(x)g^{\prime }(x)$

quotient rule

$\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right]=\frac{f^{\prime }(x)g(x)-f(x)g^{\prime }(x)}{g(x{)}^2}$

Hers is a subset of the quotient rule.

$\frac{d}{dx}\left[\frac{1}{f(x)}\right]=-\frac{f^{\prime }(x)}{f(x{)}^2}$

chain rule

$\frac{d}{dx}\left[f(g(x))\right]=f^{\prime }(g(x))\cdot g^{\prime }(x)$

more derivative rules

$\frac{d}{dx}\left[\sin (x)\right]=\cos (x)$

$\frac{d}{dx}\left[\cos (x)\right]=-\sin (x)$

$\frac{d}{dx}\left[e^x\right]=e^x$

$\frac{d}{dx}\left[\ln (x)\right]=\frac{1}{x}=x^{-1}$

$\frac{d}{dx}\left[\tan (x)\right]=\frac{1}{{\cos }^2(x)}={\sec }^2(x)$

$\frac{d}{dx}\left[\cot (x)\right]=\frac{-1}{{\sin }^2(x)}=-{\csc }^2(x)$

$\frac{d}{dx}\left[\sec (x)\right]=\frac{\sin (x)}{{\cos }^2(x)}=\tan (x)\ast \sec (x)$

$\frac{d}{dx}\left[\csc (x)\right]=\frac{-\cos (x)}{{\sin }^2(x)}=-\cot (x)\ast \csc (x)$

$\text{if }a>0\text{, }\frac{d}{dx}\left[a^x\right]=\ln (a)\ast a^x$

$\text{if }a>0\text{, and }a\ne 1\text{, }\frac{d}{dx}\left[{\log }_a(x)\right]=\frac{1}{\ln (a)\ast x}$

$f(x)=x^5+2x^3-x^2$

$f^{\prime }(x)=\frac{d}{dx}\left[f(x)\right]=\frac{d}{dx}\left[x^5+2x^3-x^2\right]$

$f^{\prime }(x)=\frac{d}{dx}\left[x^5\right]+\frac{d}{dx}\left[2x^3\right]-\frac{d}{dx}\left[x^2\right]$

$f^{\prime }(x)=5x^4+6x^2-2x^1$

implicit differentiation

we have this very hard equation:

$x^2+(y-x{)}^3=28$

the graph of this equation is continuous, but the slope changes dramatically at different x values.

take the derivative of this equation:

$\frac{d}{dx}\left[x^2+(y-x{)}^3\right]=\frac{d}{dx}\left[28\right]$

apply derivative rules:

$2x+3(y-x{)}^2(\frac{dy}{dx}-1)=0$

$2x-3(y-x{)}^2+3(y-x{)}^2\cdot \frac{dy}{dx}=0$

$3(y-x{)}^2\cdot \frac{dy}{dx}=3(y-x{)}^2-2x$

$\frac{dy}{dx}=\frac{3(y-x{)}^2-2x}{3(y-x{)}^2}$

okay, realize how this derivative equation has x and y in it. this is what makes it an implicit differentiation.

this video is an example of how implicit differs from using explicit differentiation
https://www.khanacademy.org/math/ap-calculus-ab/ab-differentiation-2-new/ab-3-2/v/showing-explicit-and-implicit-differentiation-give-same-result

derivatives of inverse functions

the inverse of $f(x)$ is $f^{-1}(x)$.

the derivative of $f(x)$ is $f^{\prime }(x)$.

$f^{\prime }(x)=\frac{1}{f^{-1}(f(x))}$

this is really exciting! also,

$\frac{d}{dx}\left[{\sin }^{-1}(x)\right]=\frac{1}{\sqrt{1-x^2}}$

$\frac{d}{dx}\left[{\cos }^{-1}(x)\right]=-\frac{1}{\sqrt{1-x^2}}$

$\frac{d}{dx}\left[{\tan }^{-1}(x)\right]=\frac{1}{1+x^2}$

second derivatives

we can take the derivative of the derivative of a function. this is called the second derivative. the second derivative of $f$ is $\frac{d}{dx}\left[\frac{d}{dx}\left[f\right]\right]$, or $f^{\prime \prime }$

hidden derivatives

what is $\underset{x\to 0}{\lim }\frac{3(2+x{)}^4-3(2{)}^4}{h}$

this limit expression has the form:

$\underset{x\to 0}{\lim }\frac{f(t+x)-f(t)}{x}$

we can tell that $t$, or the x value, is 2.

this means that we need to evaluate $f^{\prime }(2)$

in this case, $f(x)=3x^4$, and $f^{\prime }(x)=12x^3$

so the answer is $f^{\prime }(2)$, which is $12(2{)}^3$, which is $96$

related derivatives

https://www.khanacademy.org/math/ap-calculus-ab/ab-diff-contextual-applications-new/ab-4-5/v/falling-ladder-related-rates
here is example of a related rates problem

the differentiable functions $f$ and $g$ are related by the following equation:

$\sin (f)+\cos (g)=\sqrt{2}$

also, $\frac{df}{dt}=5$

find $\frac{dg}{dt}$ when $y=\frac{\pi }{4}$ and $0<x<\frac{\pi }{2}$

$\frac{d}{dt}\left[\sin (f(t))\right]+\frac{d}{dt}\left[\cos (g(t))\right]=\frac{d}{dt}\left[\sqrt{2}\right]$

$\frac{d}{dt}\left[\sin (f)\right]\ast \frac{df}{dt}+\frac{d}{dt}\left[\cos (g)\right]\ast \frac{dg}{dt}=0$

in this case, $\frac{df}{dt}=5$ and $y=\frac{\pi }{4}$, and after solving for $g$ we know that $g=\frac{\pi }{4}$ so we can simplify:

$5\ast \cos (\frac{\pi }{4})-\frac{dy}{dt}\ast \sin (\frac{\pi }{4})=0$

$5-\frac{dy}{dt}=0$

$\frac{dy}{dt}=5$