Question

Find the value $\sin \left( {2{{\cos }^{ - 1}}x} \right)$

Answer 1

Hint: We use the formulas of differentiation to solve this type of problem. If there is function $f\left( {g\left( x \right)} \right)$ is given the its differentiation can be given by chain rule as-
$\dfrac{d}{{dx}}\left( {f\left( {g\left( x \right)} \right)} \right) = f'\left( {g\left( x \right)} \right) \times g'\left( x \right)$

Complete step-by-step answer:
The given function is $\sin \left( {2{{\cos }^{ - 1}}x} \right)$
Differentiating the function by chain rule
$\begin{gathered}
  \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\sin (2{\cos ^{ - 1}}x) \\
   = \cos (2{\cos ^{ - 1}}x)\dfrac{d}{{dx}}2{\cos ^{ - 1}}x \\
\end{gathered} $
Again differentiating the remaining part
$\begin{gathered}
   = \cos (2{\cos ^{ - 1}}x)\left( {\dfrac{{ - 1(2)}}{{\sqrt {1 - {x^2}} }}} \right) \\
   = \dfrac{{ - (2)}}{{\sqrt {1 - {x^2}} }}\cos (2{\cos ^{ - 1}}x) \\
\end{gathered} $
Hence, the differentiation of function$\sin \left( {2{{\cos }^{ - 1}}x} \right)$is$\dfrac{{ - (2)}}{{\sqrt {1 - {x^2}} }}\cos (2{\cos ^{ - 1}}x)$

Note: Firstly we need to identify all the functions used and then we can see it is of the form $f\left( {g\left( x \right)} \right)$ SO, we will apply chain rule very carefully. Avoid calculation mistakes during solving problems and every part should be written systematically.