Question

How do find the derivative of $y = {(\sin x)^{\ln x}}?$

Answer 1

We can use implicit differentiation here. Implicit differentiation is a process to differentiate any implicit equation with respect to its variable ‘x’ and the other variables will be the function of ‘x’. First, we can find the derivative of ‘y’. The derivative of ‘y’ will then be used to find the differentiation of ${(\sin x)^{\ln x}}$.

Complete step by step answer:
The given equation is:
$y = {(\sin x)^{\ln x}}$
According to this equation, we will pull out the $\ln x$. So, first, we will take logs on both the sides of the equation. We will now put the famous rule of logarithms which is:
$\ln {x^a} = a\ln x$
When we apply this rule in our equation then we get:
$\ln y = \ln {(\sin x)^{\ln x}}$
$\Rightarrow \ln y = \ln x \cdot \ln (\sin x)$
Now, we will do differentiation on both the sides:
$\Rightarrow \dfrac{d}{{dx}}\left( {\ln y} \right) = \dfrac{d}{{dx}}\left[ {\ln \cdot \ln x(\sin x)} \right]$
We know that the derivative of $\ln y = \dfrac{1}{y}$. But the derivative of $y = \dfrac{{dy}}{{dx}}$

Now, the left side of our equation also called as LHS is:
$\dfrac{{dy}}{{dx}} \times \dfrac{1}{y}$
And the right side of the part is:
$\dfrac{d}{{dx}}\left[ {\ln \cdot \ln x(\sin x)} \right]$
We have to differentiate the right side. By applying the Chain Rule and the Product Rule of differentiation, we get the right-side part as:
$= \left( {\ln (\sin x) \cdot \dfrac{1}{x}} \right) + \left( {\ln x \cdot \dfrac{1}{{\sin x}} \cdot \cos x} \right)$
Now, our equation looks like:
$\Rightarrow \dfrac{1}{y} \cdot \dfrac{{dy}}{{dx}} = \left( {\ln (\sin x) \cdot \dfrac{1}{x}} \right) + \left( {\ln x \cdot \dfrac{1}{{\sin x}} \cdot \cos x} \right)$
When we open the bracket part, we get:
$\Rightarrow \dfrac{1}{y} \cdot \dfrac{{dy}}{{dx}} = \dfrac{{\ln (\sin x)}}{x} + \dfrac{{\ln x \cdot \cos x}}{{\sin x}}$

Now, either we can shift $y$to the other side. If we do this, then it will get multiplied with the whole term over there. Otherwise, we can multiply both the sides with $y$. We will get the same answer. So, after multiplying $y$on both the sides we get:
$\Rightarrow \dfrac{1}{y} \cdot y \cdot \dfrac{{dy}}{{dx}} = y \cdot \left( {\dfrac{{\ln (\sin x)}}{x} + \dfrac{{\ln x \cdot \cos x}}{{\sin x}}} \right)$
$\Rightarrow \dfrac{{dy}}{{dx}} = y \cdot \left( {\dfrac{{\ln (\sin x)}}{x} + \dfrac{{\ln x \cdot \cos x}}{{\sin x}}} \right)$
Now, to get back the answer in the terms of ‘x’, we will use our first equation here. Our first equation was:
$y = {(\sin x)^{\ln x}}$
When we put the value of ‘y’ in the above equation then we get:
$\therefore \dfrac{{dy}}{{dx}} = {(\sin x)^{\ln x}}\left( {\dfrac{{\ln (\sin x)}}{x} + \dfrac{{\ln x \cdot \cos x}}{{\sin x}}} \right)$
Therefore, this is our final answer.

Note: We cannot take the derivative of $y$simply as $1$ here. Yes, when we apply the Chain Rule of differentiation to find the derivative of $y$, then we get the answer as $1$, but here we cannot use it. Do the differentiation using the chain rule and the product rule correctly.