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

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

Explanation

Solution

Hint : We use the formulas of differentiation to solve this type of problem. If there is function f(g(x))f\left( {g\left( x \right)} \right) is given the its differentiation can be given by chain rule as-
ddx(f(g(x)))=f′(g(x))×g′(x)\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⁡(2cos⁡−1x)\sin \left( {2{{\cos }^{ - 1}}x} \right)
Differentiating the function by chain rule
dydx=ddxsin⁡(2cos⁡−1x) =cos⁡(2cos⁡−1x)ddx2cos⁡−1x \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
=cos⁡(2cos⁡−1x)(−1(2)1−x2) =−(2)1−x2cos⁡(2cos⁡−1x) \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 functionsin⁡(2cos⁡−1x)\sin \left( {2{{\cos }^{ - 1}}x} \right)is−(2)1−x2cos⁡(2cos⁡−1x)\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(g(x))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.