Question

Differentiate the function w.r.t x.
 ${x^{x\cos x}} + \dfrac{{{x^2} + 1}}{{{x^2} - 1}}$ .

Answer 1

Hint: In this question remember to differentiate each function separately and use rules such as product rule which is given as; $\dfrac{{d\left( {pqr} \right)}}{{dx}} = qr\dfrac{{dp}}{{dx}} + rp\dfrac{{dq}}{{dx}} + qp\dfrac{{dr}}{{dx}}$ and 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), using this information will help you to approach the solution.

Complete step-by-step answer:
According to the given information we have function ${x^{x\cos x}} + \dfrac{{{x^2} + 1}}{{{x^2} - 1}}$
Let$y = {x^{x\cos x}} + \dfrac{{{x^2} + 1}}{{{x^2} - 1}}$
And let $u = {x^{x\cos x}}$ and $v = \dfrac{{{x^2} + 1}}{{{x^2} - 1}}$
Now y = u + v
Differentiating both side with respect to x we get
$\dfrac{{dy}}{{dx}} = \dfrac{{du}}{{dx}} + \dfrac{{dv}}{{dx}}$ (equation 1)
\[u = {x^{x\cos x}}\]
$ \Rightarrow $\[\log u = \log ({x^{x\cos x}})\]
$ \Rightarrow $\[\log u = x\cos x\log x\]
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 $\dfrac{{d\left( {\log x} \right)}}{{dx}} = \dfrac{1}{x}$and $\dfrac{d}{{dx}}\left( {{x^2} \pm 1} \right) = 2x$
Therefore, $\dfrac{1}{v}\dfrac{{dv}}{{dx}} = \dfrac{{2x}}{{{x^2} + 1}} - \dfrac{{2x}}{{{x^2} - 1}}$
$\dfrac{{dv}}{{dx}} = v\left[ {\dfrac{{2x({x^2} - 1) - 2x({x^2} + 1)}}{{({x^2} + 1)({x^2} - 1)}}} \right]$
$\dfrac{{dv}}{{dx}} = \dfrac{{{x^2} + 1}}{{{x^2} - 1}} \times \left[ {\dfrac{{ - 4x}}{{({x^2} + 1)({x^2} - 1)}}} \right]$
$\dfrac{{dv}}{{dx}} = \dfrac{{ - 4x}}{{{{({x^2} - 1)}^2}}}$(equation 3)
Now substituting the values form (2) and (3) in the equation 1 er get
$\dfrac{{dy}}{{dx}} = {x^{x\cos x}}\left[ {\cos x(1 + \log x) - x\sin x\log x} \right] - \dfrac{{4x}}{{{{({x^2} - 1)}^2}}}$
Therefore, after differentiating this function ${x^{x\cos x}} + \dfrac{{{x^2} + 1}}{{{x^2} - 1}}$with respect to x we get ${x^{x\cos x}}\left[ {\cos x(1 + \log x) - x\sin x\log x} \right] - \dfrac{{4x}}{{{{({x^2} - 1)}^2}}}$.

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.