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$

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

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Mathematics is the most important tool of science. The quest to understand the world and the universe using mathematics is as old as civilisation, and has led to the science and technology of today. Learn about the techniques and history of mathematics on ScienceLibrary.info.

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Gregor Mendel is known as 'the Father of Modern Genetics' published his work, *Versuche über Pflanzenhybriden* [Experiments on Plant Hybridization], on hereditary traits of peas, in 1866. The insight of his work became apparent later when the science of genetics confirmed his findings.

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