Question

Differentiate $w.r.$ to $x$
${\sin }^{-1}\left(\frac{a\cos x+b\sin x}{\sqrt{a^2+b^2}}\right)$

Answer 1

Let $y={\sin }^{-1}\left(\frac{a\cos x+b\sin x}{\sqrt{a^2+b^2}}\right)$

$\Rightarrow \sin y=\frac{a\cos x+b\sin x}{\sqrt{a^2+b^2}}$

$\Rightarrow \cos y\frac{dy}{dx}=\frac{1}{\sqrt{a^2+b^2}}\frac{d}{dx}(a\cos x+b\sin x)$

$\Rightarrow \cos y\frac{dy}{dx}=\frac{1}{\sqrt{a^2+b^2}}(-a\sin x+b\cos x)$

$\Rightarrow \sqrt{1-{\sin }^{2}y}\frac{dy}{dx}=\frac{-a\sin x+b\cos x}{\sqrt{a^2+b^2}}$

$\Rightarrow \sqrt{1-{\left(\frac{a\cos x+b\sin x}{\sqrt{a^2+b^2}}\right)}^2}\frac{dy}{dx}=\frac{-a\sin x+b\cos x}{\sqrt{a^2+b^2}}$

$\therefore \frac{dy}{dx}=\frac{-a\sin x+b\cos x}{\sqrt{a^2+b^2}}\times \frac{1}{\sqrt{1-{\left(\frac{a\cos x+b\sin x}{\sqrt{a^2+b^2}}\right)}^2}}$

$\frac{dy}{dx}=\frac{-a\sin x+b\cos x}{\sqrt{a^2+b^2}}\times \frac{\sqrt{a^2+b^2}}{\sqrt{a^2+b^2-a^2{\cos }^{2}x-b^2{\sin }^{2}x-2ab\sin x\cos x}}$

$\frac{dy}{dx}=\frac{-a\sin x+b\cos x}{\sqrt{a^2+b^2}}\times \frac{\sqrt{a^2+b^2}}{\sqrt{a^2-a^2{\cos }^{2}x+b^2-b^2{\sin }^{2}x-2ab\sin x\cos x}}$

$\frac{dy}{dx}=\frac{-a\sin x+b\cos x}{\sqrt{a^2-a^2{\cos }^{2}x+b^2-b^2{\sin }^{2}x-2ab\sin x\cos x}}$

$\frac{dy}{dx}=\frac{-a\sin x+b\cos x}{\sqrt{a^2(1-{\cos }^{2}x)+b^2(1-{\sin }^{2}x)-2ab\sin x\cos x}}$

$\frac{dy}{dx}=\frac{-a\sin x+b\cos x}{\sqrt{a^2{\sin }^{2}x+b^2{\cos }^{2}x-2ab\sin x\cos x}}$

$\frac{dy}{dx}=\frac{-a\sin x+b\cos x}{\sqrt{{(a\sin x-b\cos x)}^2}}$

$\frac{dy}{dx}=\frac{-a\sin x+b\cos x}{a\sin x-b\cos x}$

$\frac{dy}{dx}=\frac{-(a\sin x-b\cos x)}{a\sin x-b\cos x}$

$\therefore \frac{dy}{dx}=-1$