Find differentiation the following function w.r.t. x. (x-1+x+1) - Vidyadip

Question

Find differentiation the following function w.r.t. $x$.$
\log (\sqrt{x-1}+\sqrt{x+1})
$

✓

Answer

Suppose that
$
y=\log (\sqrt{x-1}+\sqrt{x+1})
$
now differentiate w.r.t. $x$$
\frac{d y}{d x}=\frac{1}{(\sqrt{x-1}+\sqrt{x+1})} \times \frac{d}{d x}(\sqrt{x-1}+\sqrt{x+1})
$
$
\begin{aligned}
\frac{d y}{d x} & =\frac{1}{(\sqrt{x-1}+\sqrt{x+1})} \times\left\{\frac{1}{2 \sqrt{x-1}}+\frac{1}{2 \sqrt{x+1}}\right\} \\
& =\frac{1}{(\sqrt{x-1}+\sqrt{x+1})} \times \frac{(\sqrt{x+1}+\sqrt{x-1})}{2 \sqrt{x-1} \times \sqrt{x+1}} \\
& =\frac{1}{2 \sqrt{x^2-1}}
\end{aligned}
$

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