Question

If \[y = \dfrac{{{e^{2x}}\cos x}}{{x\sin x}}\] , then \[\dfrac{{dy}}{{dx}}\] \[ = \]
A) \[\dfrac{{{e^{2x}}\left[ {(2x - 1)\cot x - x\cos e{c^2}x} \right]}}{{{x^2}}}\]
B) \[\dfrac{{{e^{2x}}\left[ {(2x + 1)\cot x - x\cos e{c^2}x} \right]}}{{{x^2}}}\]
C) \[\dfrac{{{e^{2x}}\left[ {(2x - 1)\cot x + x\cos e{c^2}x} \right]}}{{{x^2}}}\]
D) None of these

Answer 1

Hint: In this question, we have to differentiate the given function \[y = \dfrac{{{e^{2x}}\cos x}}{{x\sin x}}\] with respect to \[x\]. We will proceed with using the formula for differentiation for division of two numbers. Then we proceed step by step and solve to get the desired result. We will also use the trigonometric identity \[{\sin ^2}x + {\cos ^2}x = 1\] as required.

Formula used: We will use the formula for differentiation for division of two terms.
I.e. \[\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v\dfrac{d}{{dx}}\left( u \right) - u\dfrac{d}{{dx}}\left( v \right)}}{{{v^2}}}\]
And formula for differentiation for products of two terms.
 I.e. \[\dfrac{d}{{dx}}\left( {uv} \right) = v\dfrac{d}{{dx}}\left( u \right) + u\dfrac{d}{{dx}}\left( v \right)\]

Complete step by step answer:
Consider the given question,
We are given \[y = \dfrac{{{e^{2x}}\cos x}}{{x\sin x}}\], we have to find the value of \[\dfrac{{dy}}{{dx}}\]
I.e. We have to differentiate with respect to \[x\].
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\dfrac{{{e^{2x}}\cos x}}{{x\sin x}}} \right)\] --------(A)
Now, we use the formula for differentiation for division of two terms
\[\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v\dfrac{d}{{dx}}\left( u \right) - u\dfrac{d}{{dx}}\left( v \right)}}{{{v^2}}}\]
Compare equation (A) with the above formula, then
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{x\sin x\dfrac{d}{{dx}}\left( {{e^{2x}}\cos x} \right) - ({e^{2x}}\cos x)\dfrac{d}{{dx}}\left( {x\sin x} \right)}}{{{{(x\sin x)}^2}}}\] …………………(1)
Now we find the value of \[\dfrac{\operatorname{d} }{{dx}}({e^{2x}}\cos x)\] and \[\dfrac{\operatorname{d} }{{dx}}(x\sin x)\].
I.e. \[\dfrac{\operatorname{d} }{{dx}}({e^{2x}}\cos x) = {e^{2x}}\dfrac{d}{{dx}}(\cos x) + \cos x\dfrac{d}{{dx}}({e^{2x}})\]
\[ \Rightarrow \dfrac{\operatorname{d} }{{dx}}({e^{2x}}\cos x) = {e^{2x}}( - \sin x) + \cos x(2{e^{2x}})\]
On simplifying, we have
\[ \Rightarrow \dfrac{\operatorname{d} }{{dx}}({e^{2x}}\cos x) = {e^{2x}}(2\cos x - \sin x)\] --(2)
And \[\dfrac{\operatorname{d} }{{dx}}(x\sin x) = x\dfrac{d}{{dx}}(\sin x) + \sin x\dfrac{d}{{dx}}(x)\]
\[ \Rightarrow \dfrac{\operatorname{d} }{{dx}}(x\sin x) = x\cos x + \sin x\] --- (3)
Substituting the values from equation (2) and equation (3) in equation (1) we have,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{x\sin x\dfrac{d}{{dx}}\left( {{e^{2x}}\cos x} \right) - ({e^{2x}}\cos x)\dfrac{d}{{dx}}\left( {x\sin x} \right)}}{{{{(x\sin x)}^2}}}\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{x\sin x\left( {{e^{2x}}(2\cos x - \sin x)} \right) - ({e^{2x}}\cos x)\left( {\sin x + x\cos x} \right)}}{{{{(x\sin x)}^2}}}\]
On taking common \[{e^{2x}}\] and simplifying , we have
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{e^{2x}}\left\{ {2x\sin x\cos x - x{{\sin }^2}x - \sin x\cos x - x{{\cos }^2}x} \right\}}}{{{{(x\sin x)}^2}}}\]
Again simplifying by taking common \[\sin x\cos x\] and \[x\] , we get
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{e^{2x}}\left\{ {\sin x\cos x(2x - 1) - x({{\sin }^2}x + {{\cos }^2}x)} \right\}}}{{{x^2}{{\sin }^2}x}}\]
We know that \[{\sin ^2}x + {\cos ^2}x = 1\]thus, we have
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{e^{2x}}\left\{ {\sin x\cos x(2x - 1) - x(1)} \right\}}}{{{x^2}{{\sin }^2}x}}\]
Dividing the numerator and denominator by \[{\sin ^2}x\]we have ,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{e^{2x}}\left\{ {\dfrac{{\cos x}}{{\sin x}}(2x - 1) + x(\dfrac{1}{{{{\sin }^2}x}})} \right\}}}{{{x^2}}}\]
We know that \[\cot x = \dfrac{{\cos x}}{{\sin x}}\] and \[cose{c^2}x = \dfrac{1}{{{{\sin }^2}x}}\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{e^{2x}}\left\{ {(2x - 1)\cot x + x{{\operatorname{cosec} }^2}x} \right\}}}{{{x^2}}}\]
Hence, the option \[C\] is correct.

Note:
1. We must remember the formulas for differentiation. Some common formula for differentiation are :
\[\dfrac{d}{{dx}}(\sin x) = \cos x\]
\[\dfrac{d}{{dx}}(\cos x) = - \sin x\]
\[\dfrac{d}{{dx}}({e^x}) = {e^x}\]
\[\dfrac{d}{{dx}}(x) = 1\]
2. While differentiation of function of function, we differentiate each and write as a product. For example, we know that \[\dfrac{d}{{dx}}({e^x}) = {e^x}\] but when we have to differentiate \[{e^{2x}}\], we first differentiate \[{e^{2x}}\]with respect to \[x\] and then differentiate the power (i.e. \[2x\]) to get the final differentiation .
i.e., \[\dfrac{d}{{dx}}({e^{2x}}) = {e^{2x}}\dfrac{d}{{dx}}(2x) = 2{e^{2x}}\].
3. Differentiation of constant terms is always zero. And when a constant term is present as
 a product, we keep constant outside of differentiation .