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

Differentiate the function w.r.t. x
xx−2sin⁡  x{x^x} - {2^{\sin \,\,x}}

Explanation

Solution

Suppose the given value (xx−2sin⁡  x)\left( {{x^x} - {2^{\sin \,\,x}}} \right) into two variables. Thereafter, we will solve separately, by using differentiation of the function with respect to x{\text{x}}.

Complete step by step solution:
Lety=xx−2sin⁡  xy = {x^x} - {2^{\sin \,\,x}}
Also, let xx=u{x^x} = u and 2sin⁡x  =v{2^{\sin x}}\,\, = v
∴    y=u−v\therefore \,\,\,\,y = u - v
Differentiating both sides with respect tox{\text{x}}.
⇒    dydx=dudx−dvdx\Rightarrow \,\,\,\,\dfrac{{dy}}{{dx}} = \dfrac{{du}}{{dx}} - \dfrac{{dv}}{{dx}}
First we will solve: u=xxu = {x^x}
Taking logarithm on both sides, we obtain
log⁡ u=x log x\log \,u = x{\text{ }}log{\text{ }}x
Differentiating both sides with respect tox{\text{x}}, we obtain
1u  dudx=[ log⁡x×ddx(x)  +x×ddx(log⁡x)]\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]
⇒   dudx=  u  [log⁡x×1  +x×1xddx(x)]\Rightarrow \,\,\,\dfrac{{du}}{{dx}} = \,\,u\,\,\left[ {\log x \times 1\,\, + x \times \dfrac{1}{x}\dfrac{d}{{dx}}\left( x \right)} \right]
dudx=x2(log⁡ x+1)\dfrac{{du}}{{dx}} = {x^2}\left( {\log \,x + 1} \right)   (∵u=x2)\;\left( {\because u = {x^2}} \right)
v=2sin⁡  xv = {2^{\sin \,\,x}}
Taking logarithm on both the sides with respect to x{\text{x}}, obtain
log⁡ v = sin x log 2\log \,v{\text{ }} = {\text{ }}sin{\text{ }}x{\text{ }}log{\text{ }}2
Differentiating both sides with respect to x we obtain
1v  .  dvdx  =  log⁡ 2  .  ddx  sin⁡  x\dfrac{1}{v}\,\,.\,\,\dfrac{{dv}}{{dx}}\,\, = \,\,\log \,2\,\,.\,\,\dfrac{d}{{dx}}\,\,\sin \,\,x (∴log⁡2)\left( {\therefore \log 2} \right)is a constant term
⇒    dvdx  =  v  log⁡  2   cos⁡  x\Rightarrow \,\,\,\,\dfrac{{dv}}{{dx}}\,\, = \,\,v\,\,\log \,\,2\,\,\,\cos \,\,x
⇒    dvdx  =  2sin⁡  x  cos⁡ x   log⁡2\Rightarrow \,\,\,\,\dfrac{{dv}}{{dx}}\,\, = \,\,{2^{\sin \,\,x}}\,\,\cos \,x\,\,\,\log 2
Therefore, adding the values of dudxanddvdx\dfrac{{du}}{{dx}}and\dfrac{{dv}}{{dx}}, we will get
∴     dydx  =  xx(1+log⁡x)−2sin⁡xcos⁡x  log⁡2\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.