Answer
The derivative of $f$ at x, denoted by $f'\left( x \right)$, is defined by $\underset{h\to 0}{\mathop{\lim }}\,$ $\frac{f\left( x+h \right)-f\left( x \right)}{h}$ provided that this limit exists.
Work Step by Step
Consider the provided statement,
The derivative of $f$, denoted as $f'\left( x \right)$, is defined by $\underset{h\to 0}{\mathop{\lim }}\,$ $\frac{f\left( x+h \right)-f\left( x \right)}{h}$ provided that this limit exists.
For example,
Consider the function $f\left( x \right)=2{{x}^{2}}$
The derivative of the function $f\left( x \right)=2{{x}^{2}}$ is denoted by $f'\left( x \right)$ and is defined by,
$\begin{align}
& \underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( x+h \right)-f\left( x \right)}{h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{2{{\left( x+h \right)}^{2}}-2{{x}^{2}}}{h} \\
& =\underset{h\to 0}{\mathop{\lim }}\,\frac{2{{x}^{2}}+2{{h}^{2}}+4xh-2{{x}^{2}}}{h} \\
& =\underset{h\to 0}{\mathop{\lim }}\,\frac{2{{h}^{2}}+4xh}{h} \\
& =\underset{h\to 0}{\mathop{\lim }}\,\left( 2h+4x \right)
\end{align}$
Taking the limit inside,
$\begin{align}
& \underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( x+h \right)-f\left( x \right)}{h}=2\underset{h\to 0}{\mathop{\lim }}\,h+4x \\
& =4x
\end{align}$
Thus, $f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( x+h \right)-f\left( x \right)}{h}=4x$
Therefore, the derivative of $f$ at x, denoted by $f'\left( x \right)$, is defined by $\underset{h\to 0}{\mathop{\lim }}\,$ $\frac{f\left( x+h \right)-f\left( x \right)}{h}$ provided that this limit exists.
