Question
Answers

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

Answer Verified Verified
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.