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The limit used to define the slope of a tangent line is also used to define one of the two fundamental operations of calculus—differentiation.

The derivative of at is

f'(x) = $\frac{\text{d}y}{\text{d}x} =\lim_{\triangle x \rightarrow 0}\frac{f(x+\triangle x)-f(x)}{\triangle x}$

provided the limit exists. For all for which this limit exists, f' is a function of x

This definition of derivative of a function is used to find the instantaneous rate of change at a particular value of x. The process of finding the derivative of a function is called the

A function is differentiable at when its derivative exists at and is differentiable on an open interval (a,b) when it is differentiable at every point in the interval.

The other notation for f'(x) is 1) y' 2) $\frac{\text{d}y}{\text{d}x}$ 3) $\frac{\text{d}[f(x)]}{\text{d}x}$ 4) $D_{x}[y]$

f'(x) = $\lim_{\triangle x \rightarrow 0}\frac{f(x+\triangle x)-f(x)}{\triangle x}$

Now in the given function replace x by (x + $\triangle x$) and simplify

=$\lim_{\triangle x \rightarrow 0}\frac{[(x+\triangle x)^{3}+2(x + \triangle x)-(x^{3} + 2x)}{\triangle x}$

=$\lim_{\triangle x \rightarrow 0}\frac{(x^{3} + 3 x^{2}\triangle x + 3 x (\triangle x)^{2} + (\triangle x)^{3}+2x + 2\triangle x)-(x^{3} + 2x)}{\triangle x}$

=$\lim_{\triangle x \rightarrow 0}\frac{x^{3} + 3 x^{2}\triangle x + 3 x (\triangle x)^{2} + (\triangle x)^{3}+2x + 2\triangle x- x^{3} - 2x}{\triangle x}$

=$\lim_{\triangle x \rightarrow 0}\frac{ 3 x^{2}\triangle x + 3 x (\triangle x)^{2} + (\triangle x)^{3} + 2\triangle x}{\triangle x}$

=$\lim_{\triangle x \rightarrow 0}\frac{ \triangle x [ 3 x^{2} + 3 x (\triangle x) + (\triangle x)^{2} + 2}{\triangle x}$

=$\lim_{\triangle x \rightarrow 0} {3 x^{2} + 3 x (\triangle x) + (\triangle x)^{2} + 2}$

Plug in $ \triangle x$= 0

= $3 x^{2} + 3 x (0) + (0)^{2} + 2$

=$3 x^{2} +2 $

f'(x) = $\lim_{\triangle x \rightarrow 0}\frac{f(x+\triangle x)-f(x)}{\triangle x}$

Now in the given function replace x by (x + $\triangle x$) and simplify

=$\lim_{\triangle x \rightarrow 0}\frac{[(\sqrt{x + \triangle x})-(\sqrt{x})}{\triangle x}$

Now we will rationalize the numerator

= $\lim_{\triangle x \rightarrow 0}[\frac{(\sqrt{x + \triangle x})-(\sqrt{x})}{\triangle x}][\frac{(\sqrt{x + \triangle x})+(\sqrt{x})}{(\sqrt{x + \triangle x})+(\sqrt{x})}]$

= $\lim_{\triangle x \rightarrow 0}\frac{({x + \triangle x})-(x)}{\triangle x (\sqrt{(x + \triangle x)+ x})}$

= $\lim_{\triangle x \rightarrow 0}\frac{ \triangle x}{\triangle x (\sqrt{(x + \triangle x)+ x})}$

=$\lim_{\triangle x \rightarrow 0}\frac{ 1}{(\sqrt{(x + \triangle x)+ x})}$

Plug in $\triangle x $= 0

=$\frac{1}{\sqrt(x + 0)+\sqrt {x}}$

= $\frac{ 1}{\sqrt{x} + \sqrt{x}}$

=$\frac{ 1}{2\sqrt{x}}$

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