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Differentiate the following w.r.t $x$ : $\sin^{-1} \big[\large\frac{1}{\sqrt 2}$$\sin x+\large\frac{1}{\sqrt 2}$$\cos x\big],\large\frac{-\pi}{4}<$$x<\large\frac{\pi}{4}$.

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Toolbox:
  • $\cos A\cos B+\sin A\sin B=sin(A+B)$
  • $\large\frac{d}{dx}$$(x^n)=nx^{n-1}$
Step 1:
$y=\sin^{-1} \big[\large\frac{1}{\sqrt 2}$$\sin x+\large\frac{1}{\sqrt 2}$$\cos x\big]$
$\cos\large\frac{\pi}{4}=\large\frac{1}{\sqrt 2}$ and $\sin\large\frac{\pi}{4}=\frac{1}{\sqrt 2}$
$\;\;\;=\sin^{-1} \big[\sin\large\frac{\pi}{4}$$\sin x+\cos\large\frac{\pi}{4}$$\cos x\big]$
This can be written as
$\;\;\;=\sin^{-1} \big[\cos\large\frac{\pi}{4}$$\cos x+\sin\large\frac{\pi}{4}$$\sin x\big]$
Step 2:
But $\cos A\cos B+\sin A\sin B=\sin(A+B)$
$y=\sin^{-1}\big(\sin(\large\frac{\pi}{4}$$+x)\big)$
$y=\large\frac{\pi}{4}$$+x$
On differentiating with respect to $x$ we get,
$\large\frac{dy}{dx}$$=1$
answered Jun 28, 2013 by sreemathi.v
 

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