Question

Find the derivative of $\sin \left( {2{{\sin }^{ - 1}}x} \right)$.
A) $\dfrac{{2\cos \left( {2{{\sin }^{ - 1}}x} \right)}}{{\sqrt {1 - {x^2}} }}$
B) $\dfrac{{\cos \left( {2{{\sin }^{ - 1}}x} \right)}}{{\sqrt {1 - {x^2}} }}$
C) $\dfrac{{2\cos \left( {2{{\cos }^{ - 1}}x} \right)}}{{\sqrt {1 - {x^2}} }}$
D) $ - \dfrac{{\cos \left( {2{{\cos }^{ - 1}}x} \right)}}{{\sqrt {1 - {x^2}} }}$

Answer 1

Hint: First, let $2{\sin ^{ - 1}}x = t$ and differentiate the given function using chain rule. We will apply chain rule because $t$ is also a function of $x$. Then, use the formulas of derivatives and simplify the expression to get the required answer.

Complete step by step solution: First of we will let the expression $2{\sin ^{ - 1}}x = t$ to simplify the expression.
Hence, the expression becomes $\sin \left( t \right)$
Then, we will find the derivative of the $\sin \left( t \right)$, where $t$ is a function of $x$.
We will apply the chain rule to solve its derivative, that is we will first find the derivative of $\sin \left( t \right)$ and multiply it with the derivative of $t$.
We know that the derivative of $\sin x$ is $\cos x$.
Hence, we have,
$\dfrac{d}{{dx}}\sin \left( t \right) = \cos \left( t \right)\dfrac{d}{{dx}}\left( t \right)$
Replace the value of $t$ and differentiate it with respect to $x$.
$\dfrac{d}{{dx}}\sin \left( {2{{\sin }^{ - 1}}x} \right) = \cos \left( {2{{\sin }^{ - 1}}x} \right)\dfrac{d}{{dx}}\left( {2{{\sin }^{ - 1}}x} \right)$
Now, the derivative of ${\sin ^{ - 1}}x = \dfrac{1}{{\sqrt {1 - {x^2}} }}$
Therefore, we will get,
$
  \dfrac{d}{{dx}}\sin \left( {2{{\sin }^{ - 1}}x} \right) = \cos \left( {2{{\sin }^{ - 1}}x} \right)2\left( {\dfrac{1}{{\sqrt {1 - {x^2}} }}} \right) \\
   \Rightarrow \dfrac{d}{{dx}}\sin \left( {2{{\sin }^{ - 1}}x} \right) = \dfrac{{2\cos \left( {2{{\sin }^{ - 1}}x} \right)}}{{\sqrt {1 - {x^2}} }} \\
$

Hence, option A is the correct answer.

Note: In this question we have first assumed $2{\sin ^{ - 1}}x = t$ and then differentiated the function, followed by the differentiation of \[t\]. This is called the chain rule of derivatives. We apply the chain rule of derivatives when the function which we want to differentiate is a composite function.