Question

Differentiate the following trigonometric function \[{\sin ^2}{\text{x}}\] with respect to ${{\text{e}}^{\cos {\text{x}}}}$

Answer 1

Hint- Proceed the solution of this question first considering the given function as two different functions and then differentiate both w.r.t. x, then further dividing both outcomes will give differentiation of one function with respect to another.

Complete step-by-step answer:
Let u= \[{\sin ^2}{\text{x}}\] & v =${{\text{e}}^{\cos {\text{x}}}}$
Here, we have to Differentiate \[{\sin ^2}{\text{x}}\] with respect to ${{\text{e}}^{\cos {\text{x}}}}$.
Step1: Differentiate, u = \[{\sin ^2}{\text{x}}\] with respect to x,
The chain rule tells us how to find the derivative of a composite function.
$\dfrac{{\text{d}}}{{{\text{dx}}}}[{\text{f}}\left( {{\text{g(x)}}} \right)] = {\text{f'}}\left( {{\text{g(x)}}} \right){\text{g'(x)}}$
A function is composite if you can write it as $[{\text{f}}\left( {{\text{g(x)}}} \right)]$.
In other words, it is a function within a function, or a function of a function.
$\dfrac{{{\text{du}}}}{{{\text{dx}}}} = \dfrac{{{\text{d}}\left( {{\text{si}}{{\text{n}}^2}{\text{x}}} \right)}}{{{\text{dx}}}} = 2\sin {\text{x}}{\text{.cos x}}$
Hence, differentiation of \[{\sin ^2}{\text{x}}\] with respect to x is $\dfrac{{{\text{du}}}}{{{\text{dx}}}} = 2\sin {\text{x}}{\text{.cos x}}$
Step2 : Differentiate v = ${{\text{e}}^{\cos {\text{x}}}}$ with respect to x,
$\dfrac{{{\text{dv}}}}{{{\text{dx}}}} = \dfrac{{{\text{d}}{{\text{e}}^{{\text{cosx}}}}}}{{{\text{dx}}}} = - \sin {\text{x}}{\text{.}}{{\text{e}}^{{\text{cosx}}}}$
Hence, differentiation of ${{\text{e}}^{\cos {\text{x}}}}$ with respect to x is $\dfrac{{{\text{dv}}}}{{{\text{dx}}}} = - \sin {\text{x}}{\text{.}}{{\text{e}}^{{\text{cosx}}}}$
Step3 : Now dividing \[\dfrac{{{\text{du}}}}{{{\text{dx}}}}{\text{ by }}\dfrac{{{\text{dv}}}}{{{\text{dx}}}}\] ( we get differentiation of \[{\sin ^2}{\text{x}}\] with respect to ${{\text{e}}^{\cos {\text{x}}}}$)
i.e., \[\dfrac{{\dfrac{{{\text{du}}}}{{{\text{dx}}}}}}{{\dfrac{{{\text{dv}}}}{{{\text{dx}}}}}}{\text{ = }}\dfrac{{{\text{du}}}}{{{\text{dx}}}} \times \dfrac{{{\text{dx}}}}{{{\text{dv}}}} = \dfrac{{{\text{du}}}}{{{\text{dv}}}}\]
So on putting these values from step3 & step3
$ \Rightarrow \dfrac{{2\sin {\text{x}}{\text{.cosx}}}}{{ - \sin {\text{x}}{{\text{e}}^{{\text{cosx}}}}}}$
So \[\dfrac{{{\text{du}}}}{{{\text{dv}}}} = \]$\dfrac{{2\sin {\text{x}}{\text{.cosx}}}}{{ - \sin {\text{x}}{{\text{e}}^{{\text{cosx}}}}}}$
Hence, differentiation of \[{\sin ^2}{\text{x}}\] with respect to ${{\text{e}}^{\cos {\text{x}}}}$ is $\dfrac{{-2{\text{.cosx}}}}{{ {{\text{e}}^{{\text{cosx}}}}}}$

Note- Whenever we came up of such type of question we should understand the meaning of derivative \[\left( {\dfrac{{{\text{du}}}}{{{\text{dv}}}}} \right)\] i.e. rate of change of function u with respect to function v. Hence in the above solution what we are doing, in the first and second step we are differentiating both functions w.r.t. x then on dividing dx term got cancelled hence automatically we get the desired result i.e. rate of change of $1^{st}$ function with respect to $2^{nd}$ function.