Answer

Verified

402.6k+ views

**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.

Recently Updated Pages

The base of a right prism is a pentagon whose sides class 10 maths CBSE

A die is thrown Find the probability that the number class 10 maths CBSE

A mans age is six times the age of his son In six years class 10 maths CBSE

A started a business with Rs 21000 and is joined afterwards class 10 maths CBSE

Aasifbhai bought a refrigerator at Rs 10000 After some class 10 maths CBSE

Give a brief history of the mathematician Pythagoras class 10 maths CBSE

Trending doubts

Difference Between Plant Cell and Animal Cell

Give 10 examples for herbs , shrubs , climbers , creepers

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Name 10 Living and Non living things class 9 biology CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

Select the word that is correctly spelled a Twelveth class 10 english CBSE

Fill the blanks with proper collective nouns 1 A of class 10 english CBSE