Question

How do you find the derivatives of $y={{\left( \sin \theta \right)}^{\tan \theta }}$ by logarithmic differentiation?

Answer 1

Hint: To find the derivative of $y={{\left( \sin \theta \right)}^{\tan \theta }}$ by logarithmic differentiation, we have to first of all take log on both the sides and then differentiate both the sides with respect to $\theta $. After taking log on both sides, we have to use the property of logarithm that $\log {{a}^{b}}=b\log a$ for smoother differentiation.

Complete step by step answer:
The expression given in the above problem of which we have to do the logarithmic differentiation is:
$y={{\left( \sin \theta \right)}^{\tan \theta }}$ …………… Eq. (1)
First of all, we are going to take log on both sides of the above equation.
$\log y=\log {{\left( \sin \theta \right)}^{\tan \theta }}$ ………… Eq. (2)
We know that there is a property in logarithm which says that:
$\log {{a}^{b}}=b\log a$
On applying the above property of log in eq. (2) the exponent of $\sin \theta $ i.e. $\tan \theta $ will come before log and we get,
$\log y=\tan \theta \left( \log \sin \theta \right)$ ……… Eq. (3)
Now, differentiating both the sides of the above equation with respect to $\theta $ we get,
$\dfrac{1}{y}\left( \dfrac{dy}{d\theta } \right)=\tan \theta \left( \dfrac{1}{\sin \theta } \right)\cos \theta +\left( \log \left( \sin \theta \right) \right)\left( {{\sec }^{2}}\theta \right)$ …………. Eq. (4)
We have done the above derivative by using the derivative of $\log x$ with respect to $x$ which is equal to $\dfrac{1}{x}$ and the R.H.S of the q. (3) is differentiated by using product rule first followed by chain rule.
Now, we can write $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ in eq. (4) and further simplifying eq. (4) we get,
$\dfrac{1}{y}\left( \dfrac{dy}{d\theta } \right)=\dfrac{\sin \theta }{\cos \theta }\left( \dfrac{1}{\sin \theta } \right)\cos \theta +\left( \log \left( \sin \theta \right) \right)\left( {{\sec }^{2}}\theta \right)$ ………. Eq. (5)
In the R.H.S of the above equation, $\sin \theta \And \cos \theta $ from the numerator and denominator will be cancelled out and we are left with:
$\dfrac{1}{y}\left( \dfrac{dy}{d\theta } \right)=1+\left( \log \left( \sin \theta \right) \right)\left( {{\sec }^{2}}\theta \right)$ ………… Eq. (6)
Multiplying y on both the sides of eq. (6) we get,
$\dfrac{1}{y}\left( \dfrac{dy}{d\theta } \right)\times y=\left( 1+\left( \log \left( \sin \theta \right) \right)\left( {{\sec }^{2}}\theta \right) \right)\times y$
In the above equation, y will be cancelled out in the numerator and denominator and we are left with:
$\left( \dfrac{dy}{d\theta } \right)=\left( 1+\left( \log \left( \sin \theta \right) \right)\left( {{\sec }^{2}}\theta \right) \right)\times y$
Now, substituting $y={{\left( \sin \theta \right)}^{\tan \theta }}$ in the above equation and we get,
$\left( \dfrac{dy}{d\theta } \right)=\left( 1+\left( \log \left( \sin \theta \right) \right)\left( {{\sec }^{2}}\theta \right) \right)\times {{\left( \sin \theta \right)}^{\tan \theta }}$

Hence, we differentiated the given expression using logarithmic differentiation and the result is:
$\left( \dfrac{dy}{d\theta } \right)=\left( 1+\left( \log \left( \sin \theta \right) \right)\left( {{\sec }^{2}}\theta \right) \right)\times {{\left( \sin \theta \right)}^{\tan \theta }}$

Note: To make the derivative easier, you should know the logarithm base exponent property which is equal to:
$\log {{a}^{b}}=b\log a$
This property of logarithm will save your time in examination in differentiating the given expression otherwise you will be lost because differentiating $\log {{\left( \sin \theta \right)}^{\tan \theta }}$ is a bit complex and time consuming.
The mistake that could be possible in the above solution is you might forget to multiply “y” on both the sides in eq. (6) so make sure you won’t commit such a mistake.