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Question: Differentiate \( {x^{{{\sin }^{ - 1}}x}} \) w.r.t \( {\sin ^{ - 1}}x \) . A. \( {x^{{{\sin }^{ - 1...

Differentiate xsin⁡−1x{x^{{{\sin }^{ - 1}}x}} w.r.t sin⁡−1x{\sin ^{ - 1}}x .
A. xsin⁡−1x[log⁡x+sin⁡−1x⋅(1−x2)x]{x^{{{\sin }^{ - 1}}x}}\left[ {\log x + {{\sin }^{ - 1}}x \cdot \dfrac{{\sqrt {(1 - {x^2})} }}{x}} \right]
B. −xsin⁡−1x[log⁡x+sin⁡−1x⋅(1−x2)x]- {x^{{{\sin }^{ - 1}}x}}\left[ {\log x + {{\sin }^{ - 1}}x \cdot \dfrac{{\sqrt {(1 - {x^2})} }}{x}} \right]
C. xsin⁡−1x[log⁡x+sin⁡−1x⋅(1+x2)x]{x^{{{\sin }^{ - 1}}x}}\left[ {\log x + {{\sin }^{ - 1}}x \cdot \dfrac{{\sqrt {(1 + {x^2})} }}{x}} \right]
D. −xsin⁡−1x[log⁡x+sin⁡−1x⋅(1+x2)x]- {x^{{{\sin }^{ - 1}}x}}\left[ {\log x + {{\sin }^{ - 1}}x \cdot \dfrac{{\sqrt {(1 + {x^2})} }}{x}} \right]

Explanation

Solution

Hint : As we can see that we have to differentiate the given function with respect to sin⁡−1x{\sin ^{ - 1}}x . In this question we will assume the expressions into smaller variables to solve the easier problem. Then we take the log⁡\log to the both sides of the equation. We should know that the formula of log is that if there is log⁡(a)b\log {(a)^b} , then it can be written as blog⁡ab\log a .

Complete step by step solution:
Let us assume that y=xsin⁡−1xy = {x^{{{\sin }^{ - 1}}x}} and the power i.e. z=sin⁡−1x⇒z=1sin⁡xz = {\sin ^{ - 1}}x \Rightarrow z = \dfrac{1}{{\sin }}x . It can be written as sin⁡z=x\sin z = x .
Now we have to find dydz\dfrac{{dy}}{{dz}} . We have y=(sin⁡z)zy = {(\sin z)^z} , by taking log on the both sides we have log⁡y=log⁡(sin⁡z)z\log y = \log {(\sin z)^z} .
By applying the above logarithm formula we have: log⁡y=zlog⁡sin⁡z\log y = z\log \sin z .
Now we will differentiate it w.r.t zz (z=sin⁡−1x)(z = {\sin ^{ - 1}}x) .
We can write it as 1ydydz=log⁡sin⁡z+zcos⁡zsin⁡z\dfrac{1}{y}\dfrac{{dy}}{{dz}} = \log \sin z + z\dfrac{{\cos z}}{{\sin z}} , by transferring yy to the other side of the equation: dydz=y(log⁡sin⁡z+zcos⁡zsin⁡z)\dfrac{{dy}}{{dz}} = y\left( {\log \sin z + z\dfrac{{\cos z}}{{\sin z}}} \right) .
If we have sin⁡z=x\sin z = x , then we can write the value of cos⁡z=1−sin⁡2z=1−x2\cos z = \sqrt {1 - {{\sin }^2}z} = \sqrt {1 - {x^2}} .
By putting the values back in the equation we have: xsin⁡−1x[log⁡x+sin⁡−1x⋅(1−x2)x]{x^{{{\sin }^{ - 1}}x}}\left[ {\log x + {{\sin }^{ - 1}}x \cdot \dfrac{{\sqrt {(1 - {x^2})} }}{x}} \right] .
Hence the correct option is (a) xsin⁡−1x[log⁡x+sin⁡−1x⋅(1−x2)x]{x^{{{\sin }^{ - 1}}x}}\left[ {\log x + {{\sin }^{ - 1}}x \cdot \dfrac{{\sqrt {(1 - {x^2})} }}{x}} \right] .
So, the correct answer is “Option A”.

Note : We should note that to find the value of cos⁡z\cos z , we have used an identity which is sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1 . So to find the value of cos⁡θ=1−sin⁡2θ\cos \theta = \sqrt {1 - {{\sin }^2}\theta } . We should solve carefully to avoid any calculation mistakes and we should be fully aware of the trigonometric identities and how to differentiate them.