Question

If $y = {\left( {\sin x} \right)^x} + {x^x}$, then what is the value of $\dfrac{{dy}}{{dx}}$?

Answer 1

Hint: In this question we are given an equation which is to be differentiated. The equation is a complex equation. Such equations are solved by taking log on both sides. First, divide the RHS of the given equation into two parts. Then, solve both the equations individually by taking log on both sides. Simplify the equations and then differentiate.

Formula used: 1) $\log \left( {{a^b}} \right) = b\log a$
2) Product rule- $\dfrac{{d(uv)}}{{dx}} = u\dfrac{{d(v)}}{{dx}} + v\dfrac{{d(u)}}{{dx}}$

Complete step-by-step answer:
We are given- $y = {\left( {\sin x} \right)^x} + {x^x}$.
The type of questions which have ${\left( {{\text{function}}} \right)^{{\text{function}}}}$ are solved by taking log on both sides. But first we divide the RHS in to two parts as follows-
$u = {\left( {\sin x} \right)^x}$ and $v = {x^x}$
This gives us the equation-
$ \Rightarrow y = u + v$
Differentiating both the sides,
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{du}}{{dx}} + \dfrac{{dv}}{{dx}}$ ….. (1)
Solving part 1,
$ \Rightarrow u = {\left( {\sin x} \right)^x}$
Taking log both sides,
$ \Rightarrow \log u = \log {\left( {\sin x} \right)^x}$
Using property- $\log \left( {{a^b}} \right) = b\log a$
$ \Rightarrow \log u = x\log \left( {\sin x} \right)$
Differentiating both the sides with respect to x,
$ \Rightarrow \dfrac{{d(\log u)}}{{dx}} = \dfrac{{d[x\log (\sin x)]}}{{dx}}$
Simplify using product rule on RHS,
$ \Rightarrow \dfrac{1}{u}\dfrac{{du}}{{dx}} = x\dfrac{{d[\log (\sin x)]}}{{dx}} + \log (\sin x)\dfrac{{d(x)}}{{dx}}$
Using chain rule,
$ \Rightarrow \dfrac{1}{u}\dfrac{{du}}{{dx}} = \dfrac{{x(\cos x)}}{{\sin x}} + \log (\sin x)$
Shifting and putting $u = {\left( {\sin x} \right)^x}$
$ \Rightarrow \dfrac{{du}}{{dx}} = {\left( {\sin x} \right)^x}[x\cot x + \log (\sin x)]$ …… (2)
Solving part 2,
$ \Rightarrow v = {x^x}$
Taking log both the sides,
$ \Rightarrow \log v = \log \left( {{x^x}} \right)$
Using property- $\log \left( {{a^b}} \right) = b\log a$
$ \Rightarrow \log v = x\log x$
Differentiating both the sides with respect to x,
$ \Rightarrow \dfrac{{d(\log v)}}{{dx}} = \dfrac{{d(x\log x)}}{{dx}}$
Simplify using product rule,
$ \Rightarrow \dfrac{1}{v}\dfrac{{dv}}{{dx}} = x\dfrac{{d(\log x)}}{{dx}} + \log x\dfrac{{d(x)}}{{dx}}$
$ \Rightarrow \dfrac{1}{v}\dfrac{{dv}}{{dx}} = \dfrac{x}{x} + \log x$
$ \Rightarrow \dfrac{1}{v}\dfrac{{dv}}{{dx}} = 1 + \log x$
Shifting and putting $v = {x^x}$,
$ \Rightarrow \dfrac{{dv}}{{dx}} = {x^x}(1 + \log x)$ ….. (3)
Putting (2) and (3) in (1),
$ \Rightarrow \dfrac{{dy}}{{dx}} = {\left( {\sin x} \right)^x}[x\cot x + \log (\sin x)] + {x^x}(1 + \log x)$
Hence, this is the required answer.

Note: 1) It must be kept in mind that the given equation has to be divided into two parts before taking log. Otherwise, the resultant answer will not be the same. This is because $\log \left( {m + n} \right) \ne \log m + \log n$.
2) While differentiating $\log \left( {\sin x} \right)$, chain rule is used. It is used to differentiate composite functions. At first, a function is assumed as $x$, then it is differentiated. After the first differentiation, the function which was assumed as $x$ is differentiated. It goes on till the equation has been completely differentiated.