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# Differentiate the function w.r.t. x${x^x} - {2^{\sin \,\,x}}$

Last updated date: 21st Jun 2024
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Hint: Suppose the given value $\left( {{x^x} - {2^{\sin \,\,x}}} \right)$ into two variables. Thereafter, we will solve separately, by using differentiation of the function with respect to ${\text{x}}$.

Complete step by step solution:
Let$y = {x^x} - {2^{\sin \,\,x}}$
Also, let ${x^x} = u$ and ${2^{\sin x}}\,\, = v$
$\therefore \,\,\,\,y = u - v$
Differentiating both sides with respect to${\text{x}}$.
$\Rightarrow \,\,\,\,\dfrac{{dy}}{{dx}} = \dfrac{{du}}{{dx}} - \dfrac{{dv}}{{dx}}$
First we will solve: $u = {x^x}$
Taking logarithm on both sides, we obtain
$\log \,u = x{\text{ }}log{\text{ }}x$
Differentiating both sides with respect to${\text{x}}$, we obtain
$\dfrac{1}{u}\,\,\dfrac{{du}}{{dx}} = \left[ {\,\log x \times \dfrac{d}{{dx}}\left( x \right)\,\, + x \times \dfrac{d}{{dx}}\left( {\log x} \right)} \right]$
$\Rightarrow \,\,\,\dfrac{{du}}{{dx}} = \,\,u\,\,\left[ {\log x \times 1\,\, + x \times \dfrac{1}{x}\dfrac{d}{{dx}}\left( x \right)} \right]$
$\dfrac{{du}}{{dx}} = {x^2}\left( {\log \,x + 1} \right)$ $\;\left( {\because u = {x^2}} \right)$
$v = {2^{\sin \,\,x}}$
Taking logarithm on both the sides with respect to ${\text{x}}$, obtain
$\log \,v{\text{ }} = {\text{ }}sin{\text{ }}x{\text{ }}log{\text{ }}2$
Differentiating both sides with respect to x we obtain
$\dfrac{1}{v}\,\,.\,\,\dfrac{{dv}}{{dx}}\,\, = \,\,\log \,2\,\,.\,\,\dfrac{d}{{dx}}\,\,\sin \,\,x$ $\left( {\therefore \log 2} \right)$is a constant term
$\Rightarrow \,\,\,\,\dfrac{{dv}}{{dx}}\,\, = \,\,v\,\,\log \,\,2\,\,\,\cos \,\,x$
$\Rightarrow \,\,\,\,\dfrac{{dv}}{{dx}}\,\, = \,\,{2^{\sin \,\,x}}\,\,\cos \,x\,\,\,\log 2$
Therefore, adding the values of $\dfrac{{du}}{{dx}}and\dfrac{{dv}}{{dx}}$, we will get
$\therefore \,\,\,\,\,\dfrac{{dy}}{{dx}}\,\, = \,\,{x^x}\left( {1 + \log x} \right) - {2^{\sin x}}\cos x\,\,\log 2$

Note: To differentiate something means to take the derivative of that value. Taking the derivative of a function is the same as finding the slope at any point, so differentiating is just finding the slope.