Question

How do you differentiate $y={{\left( \sin x \right)}^{\ln x}}?$

Answer 1

Hint: (1) Find the derivative of $'y'$ firstly then find the derivative of ${{\left( \sin x \right)}^{\ln x}}$
(2) Try to shift $\ln x$ exponent and solve it by taking.
Natural $\log $ on both sides using properties of logarithms.
E.g. $\ln {{x}^{a}}=a\ln \left( x \right)$

Complete step by step solution: We know that, $y={{\left( \sin x \right)}^{\ln x}}$
So, here we can first find the derivative of $'y'$ than after that derivative of ${{\left( \sin x \right)}^{\ln x}}$
$y={{\left( \sin x \right)}^{\ln x}}$
Taking $\log $ on both sides by using the property of logarithm,
Therefore,
$\ln y=\ln {{\left( \sin x \right)}^{\ln x}}$
$\ln y=\ln x.\ln \left( \sin x \right)...(i)$
Here ${{\left( \sin x \right)}^{\ln x}}$ converted into $\ln x.\left( \sin x \right)$ by the property of logarithm.
Now,
We will derive both sides, for the left side we will have to derivate of $\ln y=\dfrac{1}{y}$
But we can’t simply say that derivative of $y$ is $1$ (using change rule) Rather we say that it is $\dfrac{dy}{dx}$
So, the left hand side of equation will be.
$\dfrac{1}{y}.\dfrac{dy}{dx}$
Now, taking derivative on right side of the equation $(i)$
Using the product and chain rule we will get,
$\dfrac{1}{y}.\dfrac{dy}{dx}=\left( \ln \left( \sin x \right).\dfrac{1}{2} \right)+\left( \ln x.\dfrac{1}{\sin x}.\cos x \right)$
$\dfrac{1}{y}.\dfrac{dy}{dx}=\dfrac{\ln \left( \sin x \right)}{x}+\dfrac{\ln x.\cos x}{\sin x}$
Multiplying above equation on both sides by $'y'$
Therefore,
$\dfrac{1}{y}.y\dfrac{dy}{dx}=y\left( \dfrac{\ln \left( \sin x \right)}{x}+\dfrac{\ln x.\cos x}{\sin x} \right)$
Here, $y$ in multiplying and division get canceled.
$\dfrac{dy}{dx}=y\left( \dfrac{\ln \left( \sin x \right)}{x}+\dfrac{\ln \cos x}{\sin x} \right)$
Hence, to get our answer in terms of $'x'$ replace $'y'$ by $\sin {{x}^{\ln x}}$ from the original function.
Our final answer will be
$\dfrac{dy}{dx}=\sin {{\left( x \right)}^{\ln x}}\left( \dfrac{\ln \left( \sin x \right)}{x}+\dfrac{\ln x.\cos x}{\sin x} \right)$

Additional Information:
(1) The chain rule tells us how to find the derivative of a composite function.
In this way we can apply chain rule.
$\dfrac{d}{dx}\left[ f(g(x)) \right]=f'\left( g\left( x \right) \right)g'\left( x \right)$
For example: $\cos \left( {{x}^{2}} \right)$
$f(x)=\cos \left( x \right)$ and $g\left( x \right)={{x}^{2}}$ then $\cos \left( {{x}^{2}} \right)=f\left( g\left( x \right) \right)$
(2) Derivative of $\sin x$ is $\cos x$ but here for the exponent component we have to use the property of logarithm.

Note:
(1) Firstly find the derivative of $'y'$ and then of the ${{\left( \sin x \right)}^{\ln x}}$
(2) Shift $\ln x$ exponent by using the property of logarithm.
(3) Use only product and chain rules.
(4) At last replace $y$ by $\sin {{x}^{\ln x}}$ from the original function.