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Differentiate the following w.r.t x: $\sin^{-1}\bigg|\large\frac{1}{\sqrt {x+1}}\bigg|$.

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  • Chain Rule : Suppose $f$ is a real value function which is a composite of two function $u$ and $v$ $(i.e$) $f=uov$ then $\large\frac{df}{dx}=\frac{dv}{dt}$$\times \large\frac{dt}{dx}$
Step 1:
Given : $\sin^{-1}\bigg|\large\frac{1}{\sqrt {x+1}}\bigg|$.
Let $u=\large\frac{1}{\sqrt{x+1}}=$$(x+1)^{\large\frac{-1}{2}}$
$\large\frac{du}{dx}=-\large\frac{1}{2}$$(x+1)^{\Large\frac{-3}{2}}$
$\qquad\qquad=\large\frac{-1}{2(x+1)^{\Large\frac{3}{2}}}$
Step 2:
$y=\sin^{-1}u$
$\large\frac{dy}{du}=\frac{1}{\sqrt{1-u^2}}$
Therefore $\large\frac{dy}{dx}=\frac{dy}{du}$$\times \large\frac{du}{dx}$
$\qquad\qquad\;\;\;\;=\large\frac{1}{\sqrt{1-u^2}}$$\times\large\frac{-1}{2(x+1)^{\Large\frac{3}{2}}}$
Step 3:
Substituting for $u$ we get,
$\large\frac{dy}{dx}=\large\frac{1}{\sqrt{1-\bigg(\Large\frac{1}{\sqrt{1+x}}\bigg)^2}}$$\times \large\frac{-1}{2(x+1)^{\Large\frac{3}{2}}}$
On simplifying we get,
$\large\frac{dy}{dx}=\large\frac{(1+x)}{\sqrt{(1+x)-1}}\times \large\frac{-1}{2(x+1)^{\Large\frac{3}{2}}}$
$\large\frac{(1+x)^1}{(1+x)^{\large\frac{3}{2}}}=\large\frac{1}{\sqrt{1+x}}$
Therefore $\large\frac{dy}{dx}=\large\frac{-1}{2\sqrt x\sqrt{1+x}}$
answered Jun 27, 2013 by sreemathi.v
 
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