Question

What is the derivative of ${\left( {\sin x} \right)^{{{\cos }^{ - 1}}x}}$?

Answer 1

Let $y = {\left( {\sin x} \right)^{{{\cos }^{ - 1}}x}}$. To find the derivative of ${\left( {\sin x} \right)^{{{\cos }^{ - 1}}x}}$, we will first take natural logarithm on both the sides then using the chain rule and product rule of differentiation we will differentiate both the side and at last we will substitute $y = {\left( {\sin x} \right)^{{{\cos }^{ - 1}}x}}$ wherever required to find the final result.

Complete step by step answer:
This problem deals with implicit differentiation. Implicit differentiation is done by differentiating an implicit equation with respect to variable say $x$ and while treating other variables as unspecified functions of $x$.
Let $y = {\left( {\sin x} \right)^{{{\cos }^{ - 1}}x}}$.
We have to find the derivative of $y = {\left( {\sin x} \right)^{{{\cos }^{ - 1}}x}}$.
For this we will take natural logarithms on both sides.
On taking natural logarithm on both the sides, we get
$\Rightarrow \ln y = \ln \left[ {{{\left( {\sin x} \right)}^{{{\cos }^{ - 1}}x}}} \right]$
Using the Power rule of logarithm i.e., ${\log _b}\left( {{m^p}} \right) = p{\log _b}m$ we can write
$\Rightarrow \ln y = \left( {{{\cos }^{ - 1}}x} \right)\left( {\ln \left( {\sin x} \right)} \right)$
Differentiating using chain rule of differentiation i.e., $\dfrac{d}{dx}\left[ {f\left( {g\left( x \right)} \right)} \right] = f \left( {{g'}\left( x \right)} \right){g'}\left( x \right)$ on left hand side and chain rule and product rule of differentiation i.e., $\dfrac{d}{dx}\left[ {f\left( x \right)g\left( x \right)} \right] = f\left( x \right){g'}\left( x \right) + g\left( x \right){f'}\left( x \right)$ on the right hand side, we get
$\Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = \left( {{{\cos }^{ - 1}}x} \right) \times \dfrac{d}{{dx}}\left( {\ln \left( {\sin x} \right)} \right) + \ln \left( {\sin x} \right) \times \dfrac{d}{{dx}}\left( {{{\cos }^{ - 1}}x} \right)$
$\Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = \left( {{{\cos }^{ - 1}}x} \right) \times \left( {\dfrac{1}{{\sin x}} \times \dfrac{d}{{dx}}\left( {\sin x} \right)} \right) + \ln \left( {\sin x} \right) \times \left( { - \dfrac{1}{{\sqrt {1 - {x^2}} }}} \right)$
$\Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = \left( {{{\cos }^{ - 1}}x} \right) \times \left( {\dfrac{{\cos x}}{{\sin x}}} \right) - \dfrac{{\ln \left( {\sin x} \right)}}{{\sqrt {1 - {x^2}} }}$
On taking LCM, we get
$\Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = \dfrac{{\sqrt {1 - {x^2}} \cot x{{\cos }^{ - 1}}x - \ln (\sin x)}}{{\sqrt {1 - {x^2}} }}$
Multiplying both the sides by $y$, we get
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{y\left( {\sqrt {1 - {x^2}} \cot x{{\cos }^{ - 1}}x - \ln (\sin x)} \right)}}{{\sqrt {1 - {x^2}} }}$
Putting $y = {\left( {\sin x} \right)^{{{\cos }^{ - 1}}x}}$, we get
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{{\left( {\sin x} \right)}^{{{\cos }^{ - 1}}x}}\left( {\sqrt {1 - {x^2}} \cot x{{\cos }^{ - 1}}x - \ln (\sin x)} \right)}}{{\sqrt {1 - {x^2}} }}$
Therefore, the derivative of ${\left( {\sin x} \right)^{{{\cos }^{ - 1}}x}}$ is $\dfrac{{{{\left( {\sin x} \right)}^{{{\cos }^{ - 1}}x}}\left( {\sqrt {1 - {x^2}} \cot x{{\cos }^{ - 1}}x - \ln (\sin x)} \right)}}{{\sqrt {1 - {x^2}} }}$.

Note:
This is a case of differentiation of implicit function i.e., we are not able to isolate the dependent variable in an equation. Both dependent and independent variables are present in this type of function. Also, note that the technique of implicit differentiation allows one to find the derivative of $y$ with respect to $x$ without having to solve the given equation for $y$.