frac{d}{dx} left( sin^{-1} left( frac{3+4x}{5 | vedclass

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(A) Let $y = \sin^{-1} \left( \frac{3}{5} \cdot \frac{1}{\sqrt{1+x^2}} + \frac{4}{5} \cdot \frac{x}{\sqrt{1+x^2}} \right)$.Let $\cos \alpha = \frac{3}

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$\frac{1}{1+x^2}$

Vedclass · Answer

(A) Let $y = \sin^{-1} \left( \frac{3}{5} \cdot \frac{1}{\sqrt{1+x^2}} + \frac{4}{5} \cdot \frac{x}{\sqrt{1+x^2}} ight)$.Let $\cos \alpha = \frac{3}{5}$,then $\sin \alpha = \frac{4}{5}$.Also,let $\sin 	heta = \frac{x}{\sqrt{1+x^2}}$,then $\cos 	heta = \frac{1}{\sqrt{1+x^2}}$.Substituting these,we get $y = \sin^{-1} (\cos \alpha \cos 	heta + \sin \alpha \sin 	heta) = \sin^{-1} (\cos(\alpha - 	heta)) = \sin^{-1} (\sin(\frac{\pi}{2} - (\alpha - 	heta))) = \frac{\pi}{2} - \alpha + 	heta$.Since $	heta = 	an^{-1} x$,we have $y = \frac{\pi}{2} - \alpha + 	an^{-1} x$.Differentiating with respect to $x$,we get $\frac{dy}{dx} = 0 - 0 + \frac{1}{1+x^2} = \frac{1}{1+x^2}$.

$\frac{d}{dx} \left( \sin^{-1} \left( \frac{3+4x}{5\sqrt{1+x^2}} \right) \right) =$

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