Answer
See the explanation below.
Work Step by Step
Consider the function $f\left( x \right)={{x}^{n}}$ ,
Find the derivative of the function $f\left( x \right)={{x}^{n}}$ using the formula ${f}'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( x+h \right)-f\left( x \right)}{h}$ ,
${f}'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{{{\left( x+h \right)}^{n}}-{{x}^{n}}}{h}$
Now, expand ${{\left( x+h \right)}^{n}}$ by the binomial formula,
${{\left( x+h \right)}^{n}}={{x}^{n}}+n{{x}^{n-1}}h+n\left( n-1 \right){{x}^{n-2}}{{h}^{2}}+\ldots +nx{{h}^{n-1}}+{{h}^{n}}$
Now, substitute the value of ${{\left( x+h \right)}^{n}}$ in ${f}'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{{{\left( x+h \right)}^{n}}-{{x}^{n}}}{h}$ and solve further,
$\begin{align}
& {f}'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{{{x}^{n}}+n{{x}^{n-1}}h+n\left( n-1 \right){{x}^{n-2}}{{h}^{2}}+\ldots +nx{{h}^{n-1}}+{{h}^{n}}-{{x}^{n}}}{h} \\
& =\underset{h\to 0}{\mathop{\lim }}\,\frac{n{{x}^{n-1}}h+n\left( n-1 \right){{x}^{n-2}}{{h}^{2}}+\ldots +nx{{h}^{n-1}}+{{h}^{n}}}{h} \\
& =\underset{h\to 0}{\mathop{\lim }}\,\left( n{{x}^{n-1}}+n\left( n-1 \right){{x}^{n-2}}h+\ldots +nx{{h}^{n-2}}+{{h}^{n-1}} \right)
\end{align}$
Now, take the limit inside and solve further,
$\begin{align}
& {f}'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,n{{x}^{n-1}}+\underset{h\to 0}{\mathop{\lim }}\,n\left( n-1 \right){{x}^{n-2}}h+\ldots +\underset{h\to 0}{\mathop{\lim }}\,nx{{h}^{n-2}}+\underset{h\to 0}{\mathop{\lim }}\,{{h}^{n-1}} \\
& =n{{x}^{n-1}}
\end{align}$
Thus, ${f}'\left( x \right)=n{{x}^{n-1}}$
