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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.

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