Question

Find the derivative of $\sin \left(2{\sin }^{-1}x\right)$.

• A. $\frac{2\cos \left(2{\sin }^{-1}x\right)}{\sqrt{1-x^2}}$
• B. $\frac{\cos \left(2{\sin }^{-1}x\right)}{\sqrt{1-x^2}}$
• C. $\frac{2\cos \left(2{\cos }^{-1}x\right)}{\sqrt{1-x^2}}$
• D. $-\frac{\cos \left(2{\cos }^{-1}x\right)}{\sqrt{1-x^2}}$

Answer 1

The correct option is A $\frac{2\cos \left(2{\sin }^{-1}x\right)}{\sqrt{1-x^2}}$

We need to find derivative of $\sin \left(2{\sin }^{-1}x\right)$

Apply the chain rule,

Let $f=\sin \left(u\right)$

Therefore, $\frac{d}{dx}\left(\sin \left(2\arcsin \left(x\right)\right)\right)=\frac{d}{du}\left(\sin \left(u\right)\right)\frac{d}{dx}\left(2\arcsin \left(x\right)\right)$

We know that $\frac{d}{du}\left(\sin \left(u\right)\right)=\cos \left(u\right)$ and $\frac{d}{dx}\left(2\arcsin \left(x\right)\right)=2\frac{d}{dx}\left(\arcsin \left(x\right)\right)=2\cdot \frac{1}{\sqrt{1-x^2}}$

So, $\frac{d}{dx}\left(\sin \left(2\arcsin \left(x\right)\right)\right)$

$=\frac{d}{du}\left(\sin \left(u\right)\right)\frac{d}{dx}\left(2\arcsin \left(x\right)\right)$

$=\cos \left(u\right)\frac{2}{\sqrt{1-x^2}}=\cos \left(2\arcsin \left(x\right)\right)\frac{2}{\sqrt{1-x^2}}=\frac{2\cos \left(2\arcsin \left(x\right)\right)}{\sqrt{1-x^2}}$