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If 'n' is a rational number, then the function f(x)=$x^{n}$ and is differentiable and its derivative with respect to x is given by

Then, f(x + h) = $(x + h)^{n}$

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

=$\lim_{h \rightarrow 0}\frac{(x+h)^{n}-(x)^{n}}{(x + h) -(x)}$

=$\lim_{h \rightarrow 0}\frac{z^{n}-x^{n}}{ z - x }$

where z =(x +h) and z$ \rightarrow x as h \rightarrow 0 $

$\frac{\text{d}[x^{n}]}{\text{d}x}= nx^{n-1} (Since \lim_{x \rightarrow a}\frac{x^{n}-a^{n}}{x- a}= nx^{n-1})$

Find the derivative of f(x) = $x^{3}$ using the definition of the derivative.

So, f(x +h) = $(x+h)^{3}$

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

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

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

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

=$\lim_{h \rightarrow 0} [(x+h)^{2} +(x +h)(x) + x^{2}]$

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

= $x^{2} + x^{2} +x^{2}$

= $3x^{2}$

Now we will solve the same question using the power rule for derivative

$\color{red} {\frac{\text{d}[x^{n}]}{\text{d}x}= nx^{n-1}}$

$\color{red}{\frac{\text{d}[x^{3}]}{\text{d}x} = 3x^{3-1}} $

$\color{red}{\frac{\text{d}[x^{3}]}{\text{d}x} = 3x^{2}}$

2) $\frac{\text{d}[\frac{1}{x^{3}}]}{\text{d}x}=\frac{\text{d}[x^{-3}]}{\text{d}x}= -3x^{-3 -1}= -3x^{-4}$

3) $\frac{\text{d}[\sqrt{x}]}{\text{d}x} = \frac{\text{d}x^{\frac{1}{2}}}{\text{d}x}$

$=\frac{1}{2}x^{\frac{1}{2}-1}$

= $\frac{1}{2}x^{\frac{-1}{2}}$

$\frac{\text{d}[\sqrt{x}]}{\text{d}x}=\frac{1}{2\sqrt{x}}$

4) $ \frac{\text{d}[x^{1}]}{\text{d}x} = 1x^{1-1}=1x^{0}$

$ \frac{\text{d}[x^{1}]}{\text{d}x}$ = 1

5) $ \frac{\text{d}[x^{1}]}{\text{d}x} = 1x^{1-1}=1x^{0}$

6)$ \frac{\text{d}[\frac{1}{x}]}{\text{d}x} = \frac{\text{d}{x^{-1}}}{\text{d}x}$

= -1$x^{(-1-1)}$

= -1$x^{-2}$

$\frac{\text{d}[\frac{1}{x}]}{\text{d}x} =\frac{-1}{x^{2}}$

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