Question

Differentiate using chain rule $\dfrac{d}{{dx}}\left( {{e^x}\log \sin 2x} \right)$ .
A.${e^x}\left( {\log \sin 2x + 2\cot 2x} \right)$
B.${e^x}\left( {\log \cos 2x + 2\cot 2x} \right)$
C.${e^x}\left( {\log \cos 2x + \cot 2x} \right)$
D.None of these

Answer 1

First, we will use the product rule to find the differentiation of ${e^x}\log \sin 2x$. We will take $\log \sin 2x$ as the second function and ${e^x}$ as the first function. Then, we will apply the chain rule on $\log \sin 2x$ to find its differentiation. We will substitute this differentiation in the formula for product rule to find the answer.

Formulas used:
If we have to find the differentiation of a product of functions. We need to follow the product rule which states that differentiation of a product of functions is the sum of product of differentiation of 1st function with the second function and the product of differentiation of the 2nd function with 1st function.
The product rule for differentiation is given by:
1.$\left( {uv} \right)' = u'v + uv'$
2.$\dfrac{d}{{dx}}\left( {{e^x}} \right) = {e^x}$
3.$\dfrac{d}{{dx}}\left( {\log x} \right) = \dfrac{1}{x}$
4.$\dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x$
5.$\dfrac{d}{{dx}}\left( {2x} \right) = 2$
6.$\dfrac{{\cos x}}{{\sin x}} = \cot x$

Complete step-by-step answer:
We will substitute ${e^x}$ for $u$ and $\log \sin 2x$ for $v$ in the 1st formula.
$\Rightarrow$ $\dfrac{d}{{dx}}\left( {{e^x}\log \sin 2x} \right) = \dfrac{d}{{dx}}\left( {{e^x}} \right)\log \sin 2x + {e^x}\dfrac{d}{{dx}}\left( {\log \sin 2x} \right)$ .
We will use the 2nd formula to simplify the equation:
$\Rightarrow$ $\left( 1 \right){\rm{ }}\dfrac{d}{{dx}}\left( {{e^x}\log \sin 2x} \right) = {e^x}\log \sin 2x + {e^x}\dfrac{d}{{dx}}\left( {\log \sin 2x} \right)$
We know that according to the chain rule, differentiation of a function $\Rightarrow$ $h\left( x \right) = g\left( {f\left( x \right)} \right)$ will be $\dfrac{d}{{dx}}\left( {g\left( {f\left( x \right)} \right)} \right) \cdot \dfrac{d}{{dx}}\left( {f\left( x \right)} \right) \cdot \dfrac{d}{{dx}}\left( x \right)$.
We will take $h\left( x \right)$ as $\log \sin 2x$, $\log \left( x \right)$ as $g\left( x \right)$ and $\sin 2x$ as $f\left( x \right)$ and we will find differentiation of $\log \sin 2x$. We will use the 3rd, 4th, 5th and 6th formula to find differentiation of $\log \sin 2x$:
$\Rightarrow \dfrac{d}{{dx}}\left( {\log \sin 2x} \right) = \dfrac{d}{{dx}}\left( {\log \sin 2x} \right) \cdot \dfrac{d}{{dx}}\left( {\sin 2x} \right) \cdot \dfrac{d}{{dx}}\left( {2x} \right)\\\\\Rightarrow \dfrac{d}{{dx}}\left( {\log \sin 2x} \right) = \dfrac{1}{{\sin 2x}} \cdot \cos 2x \cdot 2\\\\\Rightarrow \dfrac{d}{{dx}}\left( {\log \sin 2x} \right) = \dfrac{{2\cos 2x}}{{\sin 2x}}$
We will substitute $\cot 2x$ for $\dfrac{{\cos 2x}}{{\sin 2x}}$ :
$\Rightarrow \dfrac{d}{{dx}}\left( {\log \sin 2x} \right) = 2\cot 2x$
We will substitute $2\cot 2x$ for $\dfrac{d}{{dx}}\left( {\log \sin 2x} \right)$in the 1st equation:
$\Rightarrow \dfrac{d}{{dx}}\left( {{e^x}\log \sin 2x} \right) = {e^x}\log \sin 2x + {e^x} \cdot 2\cot 2x\\\\\Rightarrow \dfrac{d}{{dx}}\left( {{e^x}\log \sin 2x} \right) = {e^x}\left( {\log \sin 2x + 2\cot 2x} \right)$
$\Rightarrow$ The differentiation of ${e^x}\log \sin 2x$ is ${e^x}\left( {\log \sin 2x + 2\cot 2x} \right)$.
Option A is the correct option.

Note: We know that any function $h\left( x \right)$ is called a composite function if it is of the form $g\left( {f\left( x \right)} \right)$. The chain rule of differentiation is used to find the derivative of such composite functions.