Differentiation of Inverse Trigonometric Functions

Example 1:

$$ y=\sin ^{-1} 3x,\;\;\; \textup{Find} \;\;\frac{dy}{dx}$$

Solution:

$$ y=\sin ^{-1} 3x$$ $$ \frac{dy}{dx}=\frac{d}{dx}\sin ^{-1} 3x$$ $$ \textup{Let}\; u= 3x$$ $$ \frac{dy}{dx}=\frac{d}{dx} \sin ^{-1} u$$ $$ \frac{dy}{dx}=\frac{d}{du} \sin ^{-1} u \frac{du}{dx}$$ $$ \because\frac{d}{dx}\sin ^{-1} x=\frac{1}{\sqrt{1-x^2}}$$ $$ \therefore\frac{dy}{dx}=\frac{1}{\sqrt{1-u^2}} \frac{d}{dx}3x$$ $$ \frac{dy}{dx}=\frac{1}{\sqrt{1-(3x)^2}}(3\frac{d}{dx}x)$$ $$ \because\frac{d}{dx} x=1 $$ $$ \therefore \frac{dy}{dx}=\frac{1}{\sqrt{1-9x^2}}(3(1))$$ $$ \frac{dy}{dx}=\frac{1}{\sqrt{1-9x^2}} (3)$$ $$ \frac{dy}{dx}=\frac{3}{\sqrt{1-9x^2}}$$