Differentiate x^2-1 x from first principle. - Vidyadip

Question

Differentiate $\frac{x^2-1}{x}$ from first principle.

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Answer

We need to find derivative of $f(x)=\frac{x^2-1}{x}$
Derivative of a function f(x) from first principle is given by
$f ^{ f }( x )=\underset{{h \rightarrow 0}}{\lim}=\frac{f(x+h)-f(x)}{h}$ \{where h is a very small positive number $\}$
$\therefore$ Derivative of $f(x)=\frac{x^2-1}{x}$ is given as
$f ^{\prime}( x )=\underset{{h \rightarrow 0}}{\lim}=\frac{f(x+h)-f(x)^x}{h}$
$\begin{array}{l}\Rightarrow f ^{\prime}( x )=\underset{{h \rightarrow 0}}{\lim}\frac{\frac{(x+h)^2-1}{x+h}-\frac{x^2-1}{\alpha}}{h} \\
\Rightarrow f ^{\prime}(x)=\underset{{h \rightarrow 0}}{\lim}\frac{\left\{(x+h)^2-1\right\} x-(x+h)\left(x^2-1\right)}{h x(x+h)} \\
\Rightarrow f ^{\prime}(x)=\underset{{h \rightarrow 0}}{\lim} \frac{\left\{(x+h)^2-1\right\} x-(x+h)\left(x^2-1\right)}{h} \times \underset{{h \rightarrow 0}}{\lim} \frac{1}{x(x+h)} \\
\Rightarrow f ^{\prime}(x)=\frac{1}{x^2} \underset{{h \rightarrow 0}}{\lim} \frac{\left\{(x+h)^2-1\right\} x-(x+h)\left(x^2-1\right)}{h} \\
\Rightarrow f^{\prime}(x)=\frac{1}{x^2} \underset{{h \rightarrow 0}}{\lim} \frac{\left\{x^2+h^2+2 x h-1\right\} x-\left\{x^3+h x^2-x-h\right\}}{h} \\
\Rightarrow f^{\prime}(x)=\frac{1}{x^2} \underset{{h \rightarrow 0}}{\lim} \frac{x^3+h^2 x+2 x^2 h-x-x^3-h x^2+x+h}{h} \\
\Rightarrow f ^{\prime}(x)=\frac{1}{x^2} \underset{{h \rightarrow 0}}{\lim} \frac{h^2 x+x^2 h+h}{h} \\
\Rightarrow f ^{\prime}(x)=\frac{1}{x^2} \underset{{h \rightarrow 0}}{\lim} \frac{h\left(h x+x^2+1\right)}{h} \\
\Rightarrow f ^{\prime}( x )=\frac{1}{x^2} \underset{{h \rightarrow 0}}{\lim}\left(h h+x^2+1\right) \\
\Rightarrow f^{\prime}(x)=\frac{1}{x^2}\left(0 \times x+x^2+1\right)=\frac{x^2+1}{x^2}=1+\frac{1}{x^2}\end{array}$
$\therefore f^{\prime}(x)=1+\frac{1}{x^2}$
Hence,
Derivative of $f(x)=\frac{x^2-1}{x^2}=1+\frac{1}{x^2}$

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