Question

Differentiate the following functions
${\sin }^{-1}\left(\frac{a+b\cos x}{b+a\cos x}\right),b>a$

Answer 1

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

Find $\frac{dy}{dx}\Rightarrow$ This can be chain rule.

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

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

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