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*As the title suggests this thread is intended as a summary of differentiation, it is by no means an attempt at an exhaustive discussion of the topic. A detailed knowledge of limits is not required but is useful, it is assumed that the reader has some knowledge of limits and how to implement them. As always I welcome comments and corrections either here of via PM.*

__The Definition of a Derivative__The derivative ([itex]f'(x)[/itex]) of a function [itex]f(x)[/itex] with respect to [itex]x[/itex] is given by the limit;

[tex]f'(x)=\lim_{h\to0}\;\frac{f(x+h)-f(x)}{h}[/tex]

Provided the limit exists. If the limit exist at a point x0 we say that the function is differentiable at this point.

As can be seen from the plots below, as [itex]h[/itex] approaches zero, the secants of the curve tend to the tangent at a point x, this gives the gradient of the curve at point x (provided the function is differentiable at point x). Note however, that in the above limit we cannot simply set [itex]h[/itex] as zero directly as our limit would [itex]\to\pm\infty[/itex] and thus, the limit would not exist. We must therefore manipulate the limit into a form where this does not occur.

http://upload.wikimedia.org/wikipedia/en/thumb/a/aa/Lim-secant.png/300px-Lim-secant.png http://upload.wikimedia.org/wikipedia/en/thumb/3/37/Tangent-calculus.png/300px-Tangent-calculus.png

Images taken from Wikipedia

__Example__Find the derivative of [itex]f(x)= x^2+2[/itex]

*Solution;*[tex]f'(x)=\lim_{h\to0}\;\frac{f(x+h)-f(x)}{h}\;\;\rightarrow \lim_{h\to0}\;\frac{(x+h)^2+2-(x^2+2)}{h}[/tex]

[tex]=\lim_{h\to0}\;\frac{\not{x^2}+h^2+2hx +\not{2}-\not{x^2}-\not{2}}{h}= lim_{h\to0}\;\frac{\not{h}(h+2x)}{\not{h}}[/tex]

[tex]\therefore f'(x)=2x[/tex]

__Finding the Gradient and Tangent to a Curve at a Point__We already now that the derivative of a function gives the gradient of a curve at any point that is differentiable. This allows us to find the gradient of the function at any point x where the curve is differentiable. This then allows us to find the equation of the tangent to the curve at any point x where the function is differentiable.

__Example__Find the equation of the tangent to the curve with equation [itex]y=f(x)=x^2+2[/itex] at x=3.

*Solution;*We know that the tangent is a straight line and therefore has an equation of the form [itex]y=mx+c[/itex]. We have already found the derivative above which gives the gradient (m) of any point x on the curve y=f(x) (provided f(x) is differentiable at x). Therefore we can say that;

[tex]m=\left. \frac{dy}{dx}\;\;\right|_{x=3}=2\cdot3=6[/tex]

We can find the corresponding y value to x=3 using the original equation [itex]y=x^2+2[/itex];

[tex]y=3^2+2=11[/tex]

We can then use this information to find the equation of the tangent at point [itex](x,y)=(3,11)[/itex];

[tex]y=mx+c \rightarrow 11=3\cdot 6 + C \Rightarrow C = -7[/tex]

[itex]\therefore[/itex] the equation of the tangent to the curve [itex]y=x^2+2[/itex] at the point [itex](x,y)=(3,11)[/itex] is [itex]y=6x-7[/itex].

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