# Chain rule

A composite function \$y = f(g(x))\$ may be differentiated by using the chain rule:

for example, let \$y = (3x - 2)^3\$.

Let \$g(x) = (3x - 2)\$, and \$f(x) = x^3\$.

Let \$g(x) = u\$, then \$y = u^3\$.

\$\${dy}/{dx} = {dy}/{du}⋅{du}/{dx}\$\$

\${dy}/{du} = 3u^2\$, and \${du}/{dx} = 3\$, so \${dy}/{dx} = {dy}/{du}⋅{du}/{dx} = 3u^2 ⋅ 3 = 9u^2\$

∴ \${dy}/{dx} = 3u^2 = 3(3x - 2)^2\$

The chain rule can also be written as: \$(f o g)'(x) = f'(g(x))⋅ g'(x)\$.

### Derivatives of Odd and Even Functions

The chain rule can be used to demonstrate that the derivative of an odd function is an even function.

If \$f(-x)=-f(x)\$, then the function \$f\$ is odd.

Therefore, \$-f'(x) = f'(x)(-1)\$, and so \$f'(-x)=f'(x)\$, confirming that \$f'\$ is an even function.

And similarly, if \$f\$ is an even function, its derivative is odd.

If \$f\$ is even, \$f(-x)=f(x)\$, and therefore \$f'(-x)(-1)=f'(x)\$. therefore, \$f'(x)=-f'(x)\$, confirming that \$f'\$ is an odd function.

Maria Agnesi, 1718 - 1799, was an Italian mathematician, who developed a curve orginally discovered by Fermat and Grandi. The series of functions became known as the witches of Agnesi, since their shapes recall witches' hats, and not because of what she might have had brewing in the kitchen.

The witches of Agnesi are of the form \$y = {a^3}/{x^2 + a^2}\$.

The simplest function, \$y = 1/(x^2+1)\$, can be differentiated by first converting it to a form with rational exponents:

\$f(x) = (x^2+1)^{-1}\$

Using the chain rule: \$f'(x) = 2x⋅(-1(x^2+1)^{-2}) = -{2x}/{(x^2+1)^{2}}\$

# Product rule

If a function is a product of two functions, the product rule may be used to find the differential.

If \$f(x) = u(x)⋅ v(x)\$, then:

\$\$f'(x) = u(x)v'(x) + u'(x)v(x) \$\$

For example, if \$f(x) = (3x + 2)(2x^2 - 1)\$ then:

\$f'(x) = (3x + 2)4x + 3(2x^2 - 1) \$

\$ = 12x^2 + 8x + 6x'2 - 3 \$

\$ = 18x^2 + 8x -3\$

Alternatively, if \$y = uv\$, then :

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

### Derivation of Product rule

Let \$f(x)=u(x)v(x)\$, where \$u\$ and \$v\$ are differentiable functions.

\$f'(x)={lim}↙{h→0}{u(x+h)v(x+h) -u(x)v(x)}\$

\$={lim}↙{h→0}{u(x+h)v(x+h) -[u(x+h)v(x) -u(x+h)v(x)] -u(x)v(x)}/h\$

\$={lim}↙{h→0}[u(x+h){v(x+h)-v(x)}/h + v(x){u(x+h)-u(x)}/h]\$

\$={lim}↙{h→0}u(x+h)⋅{lim}↙{h→0}{v(x+h)-v(x)}/h \$
\$ + {lim}↙{h→0}v(x)⋅{lim}↙{h→0}{u(x+h)-u(x)}/h\$

\$=u(x)v'(x)+v(x)u'(x)\$

## Second derivative

If \$f(x) = uv\$, then:

\$f'(x) = u'v + uv'\$

Then it follows that:

\$f″(x) = (u'v)' + (uv')' = (u″v+u'v')+(u'v'+uv″) = u″v+2u'v'+uv″\$

The general solution to \$f^n(x)\$ follows the binomial theorem and is called Leibniz's formula.

# Quotient rule

If \$y = u/v\$, \$v(x) ≠ 0\$, then:

\$\${dy}/{dx} = {v(x)u'(x) - v'(x)u(x)}/{(v(x))^2}\$\$

For example, \$y = {x^2 - 1}/{x + 1} = {(x + 1)(x - 1)}/{(x + 1)} = x - 1\$. \$f'(x) = 1\$.

Confirming with the quotient rule: \$u(x) = {x^2 - 1}\$, so \$u'(x) = 2x\$, and \$v(x) = {x + 1}\$, so \$v'(x) = 1\$

\${dy}/{dx} = {v(x)u'(x) - v'(x)u(x)}/{(v(x))^2} = {(x + 1)⋅2x - 1⋅(x^2 - 1)}/{(x + 1)^2} = {(x^2 + 2x + 1)}/{(x + 1)^2} = 1\$

## Implicit functions

An explict function is one where \$y= a_nx^n + a_{n-1}x^{n-1} +.... + a_1x + a_0\$

An implicit function is one such as a circle \$x^2 + y^2 = 1\$, i which y is not explicitly defined in terms of x.

To differentiate an implicit function, differentiate the part with \$y\$ with respect to \$y\$, then multiply this by \${dy}/{dx}\$, according to the chain and product rules.

e.g. \$x^3 + y^3 -2xy^2 = 1\$

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

\$3x^2 + 3y^2{dy}/{dx} -2y^2 - 4xy{dy}/{dx} =0\$

\$ (3y^2-4xy){dy}/{dx} =2y^2- 3x^2\$

\${dy}/{dx} = {2y^2- 3x^2}/{3y^2-4xy}\$

Content © Andrew Bone. All rights reserved. Created : January 23, 2015 Last updated :November 6, 2015

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