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Differentiate the following w.r.t $x$ : $log\bigg(x+\sqrt{ x^2+a}\bigg)$.

1 Answer

Toolbox:
  • Chain rule : $\large\frac{dy}{dx}=\frac{dy}{du}$$\times \large\frac{du}{dx}$
  • $\large\frac{d}{dx}$$(\log x)=\large\frac{1}{x}$
  • $\large\frac{d}{dx}$$(x^n)=nx^{n-1}$
Step 1:
Given : $y=\log(x+\sqrt {x^2+m})$
Let $(x+\sqrt{x^2+a})=u$
Therefore $y=\log u$
$\large\frac{dy}{du}=\large\frac{1}{u}$
$u=x+\sqrt{x^2+a}$
$\large\frac{du}{dx}$$=1+\large\frac{1}{2}$$(x^2+a)^{\large\frac{-1}{2}}(2x)$
$\quad\;\;=1+\large\frac{x}{\sqrt{x^2+a}}$
Step 2:
Therefore $\large\frac{dy}{dx}=\frac{dy}{du}$$\times \large\frac{du}{dx}$
$\large\frac{dy}{dx}=\large\frac{1}{u}$$\times \big(1+\large\frac{x}{\sqrt{x^2+a}}\big)$
Substituting for $u$,
$\large\frac{dy}{dx}=\large\frac{1}{x+\sqrt{x^2+a}}$$\times \big(\large\frac{\sqrt{x^2+a}+x}{\sqrt{x^2+a}}\big)$
$\quad\;\;=\large\frac{1}{\sqrt{x^2+a}}$
answered Jun 27, 2013 by sreemathi.v
 
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