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Question: Differentiate the function w.r.t x. \({x^{x\cos x}} + \dfrac{{{x^2} + 1}}{{{x^2} - 1}}\) ....

Differentiate the function w.r.t x.
xxcos⁡x+x2+1x2−1{x^{x\cos x}} + \dfrac{{{x^2} + 1}}{{{x^2} - 1}} .

Explanation

Solution

In this question remember to differentiate each function separately and use rules such as product rule which is given as; d(pqr)dx=qrdpdx+rpdqdx+qpdrdx\dfrac{{d\left( {pqr} \right)}}{{dx}} = qr\dfrac{{dp}}{{dx}} + rp\dfrac{{dq}}{{dx}} + qp\dfrac{{dr}}{{dx}} and chain rule i.e. ddx[f(g(x))]=f′(g(x))×g′(x)\dfrac{d}{{dx}}\left[ {f\left( {g\left( x \right)} \right)} \right] = f'\left( {g\left( x \right)} \right) \times g'\left( x \right)(here f and g are two different function), using this information will help you to approach the solution.

Complete step-by-step answer:
According to the given information we have function xxcos⁡x+x2+1x2−1{x^{x\cos x}} + \dfrac{{{x^2} + 1}}{{{x^2} - 1}}
Lety=xxcos⁡x+x2+1x2−1y = {x^{x\cos x}} + \dfrac{{{x^2} + 1}}{{{x^2} - 1}}
And let u=xxcos⁡xu = {x^{x\cos x}} and v=x2+1x2−1v = \dfrac{{{x^2} + 1}}{{{x^2} - 1}}
Now y = u + v
Differentiating both side with respect to x we get
dydx=dudx+dvdx\dfrac{{dy}}{{dx}} = \dfrac{{du}}{{dx}} + \dfrac{{dv}}{{dx}} (equation 1)
u=xxcos⁡xu = {x^{x\cos x}}
\Rightarrow $$$\log u = \log ({x^{x\cos x}})$$ \Rightarrow $$$\log u = x\cos x\log xDifferentiatingbothsideswithrespecttox,weobtain Differentiating both sides with respect to x, we obtain \dfrac{{d\left( {\log u} \right)}}{{dx}} = \dfrac{{d\left( {x\cos x\log x} \right)}}{{dx}} After applying the product rule which is given as; $\dfrac{{d\left( {pqr} \right)}}{{dx}} = qr\dfrac{{dp}}{{dx}} + rp\dfrac{{dq}}{{dx}} + qp\dfrac{{dr}}{{dx}}$ $\dfrac{1}{u}\dfrac{{du}}{{dx}} = \dfrac{d}{{dx}}(x).\cos x.\log x + x\dfrac{d}{{dx}}(\cos x).\log x + x\cos x.\dfrac{d}{{dx}}(\log x)$ We know that $\dfrac{{d\left( x \right)}}{{dx}} = 1$, $\dfrac{d}{{dx}}\left( {\cos x} \right) = - \sin x$and $\dfrac{{d\left( {\log x} \right)}}{{dx}} = \dfrac{1}{x}$ $ \Rightarrow $ $\dfrac{{du}}{{dx}} = u\left[ {1.\cos x.\log x + x.( - \sin x)\log x + x\cos x.\dfrac{1}{x}} \right]$ $ \Rightarrow $ $\dfrac{{du}}{{dx}} = {x^{x\cos x}}(\cos x.\log x - x.\sin x.\log x + \cos x)$ $ \Rightarrow $ $\dfrac{{du}}{{dx}} = {x^{x\cos x}}\left[ {\cos x(1 + \log x) - x\sin x\log x} \right]$(equation 2) $v = \dfrac{{{x^2} + 1}}{{{x^2} - 1}}$ Applying both sides, we get $\log v = \log \left( {\dfrac{{{x^2} + 1}}{{{x^2} - 1}}} \right)$ Applying the quotient rule in the above equation i.e. $\log \left( {\dfrac{x}{y}} \right) = \log x - \log y$we get $\log v = \log ({x^2} + 1) - \log ({x^2} - 1)$ Differentiating both sides with respect to x, we obtain \dfrac{1}{v}\dfrac{{dv}}{{dx}} = \dfrac{{d\left( {\log ({x^2} + 1)} \right)}}{{dx}} - \dfrac{{d\left( {\log ({x^2} - 1)} \right)}}{{dx}} By applying the chain rule i.e. $\dfrac{d}{{dx}}\left[ {f\left( {g\left( x \right)} \right)} \right] = f'\left( {g\left( x \right)} \right) \times g'\left( x \right)$(here f and g are two different function) in the above equation we get \dfrac{1}{v}\dfrac{{dv}}{{dx}} = \left( {\dfrac{{d\left( {\log ({x^2} + 1)} \right)}}{{dx}}\dfrac{{d\left( {{x^2} + 1} \right)}}{{dx}}} \right) - \left( {\dfrac{{d\left( {\log ({x^2} - 1)} \right)}}{{dx}}\dfrac{{d\left( {{x^2} - 1} \right)}}{{dx}}} \right)$$
We know that d(log⁡x)dx=1x\dfrac{{d\left( {\log x} \right)}}{{dx}} = \dfrac{1}{x}and ddx(x2±1)=2x\dfrac{d}{{dx}}\left( {{x^2} \pm 1} \right) = 2x
Therefore, 1vdvdx=2xx2+1−2xx2−1\dfrac{1}{v}\dfrac{{dv}}{{dx}} = \dfrac{{2x}}{{{x^2} + 1}} - \dfrac{{2x}}{{{x^2} - 1}}
dvdx=v[2x(x2−1)−2x(x2+1)(x2+1)(x2−1)]\dfrac{{dv}}{{dx}} = v\left[ {\dfrac{{2x({x^2} - 1) - 2x({x^2} + 1)}}{{({x^2} + 1)({x^2} - 1)}}} \right]
dvdx=x2+1x2−1×[−4x(x2+1)(x2−1)]\dfrac{{dv}}{{dx}} = \dfrac{{{x^2} + 1}}{{{x^2} - 1}} \times \left[ {\dfrac{{ - 4x}}{{({x^2} + 1)({x^2} - 1)}}} \right]
dvdx=−4x(x2−1)2\dfrac{{dv}}{{dx}} = \dfrac{{ - 4x}}{{{{({x^2} - 1)}^2}}}(equation 3)
Now substituting the values form (2) and (3) in the equation 1 er get
dydx=xxcos⁡x[cos⁡x(1+log⁡x)−xsin⁡xlog⁡x]−4x(x2−1)2\dfrac{{dy}}{{dx}} = {x^{x\cos x}}\left[ {\cos x(1 + \log x) - x\sin x\log x} \right] - \dfrac{{4x}}{{{{({x^2} - 1)}^2}}}
Therefore, after differentiating this function xxcos⁡x+x2+1x2−1{x^{x\cos x}} + \dfrac{{{x^2} + 1}}{{{x^2} - 1}}with respect to x we get xxcos⁡x[cos⁡x(1+log⁡x)−xsin⁡xlog⁡x]−4x(x2−1)2{x^{x\cos x}}\left[ {\cos x(1 + \log x) - x\sin x\log x} \right] - \dfrac{{4x}}{{{{({x^2} - 1)}^2}}}.

Note: In the above solution we came across the term “function” which can be explained as relation between the provided inputs and the outputs of the given inputs such that each input is directly related to the one output. The representation of a function is given by supposing if there is a function “f” that belongs from X to Y. Examples of functions are logarithmic functions, bijective functions, trigonometric functions, binary functions, etc.