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: $(f o g)'(x) = f'(g(x))⋅ g'(x)$.

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: $f'(x) = 2x⋅(-1(x^2+1)^{-2}) = -{2x}/{(x^2+1)^{2}}$

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 :

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

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

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.

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 $v'(x) = 1$

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

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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Satyendra Nath Bose was an Indian polymath, best known for his Physics work with Einstein, and after whom the boson is named.

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