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**Hint:**We differentiate the term in the question by assuming the whole term as a variable and then taking log on both sides of the equation. Using the property of log we open RHS and then differentiate both sides.

*If m, n are two integers then, \[\log {(m)^n} = n(\log m)\]

**Complete step-by-step answer:**

Let us assume \[y = {(\cos x)^{\cos x}}\]

Then taking log on both sides of the equation we can write.

\[\log (y) = \log [{(\cos x)^{\cos x}}]\] … (1)

Since we know the property of log, if m, n are two integers then, \[\log {(m)^n} = n(\log m)\]

Here \[m = \cos x,n = \cos x\]

Therefore, we can write \[\log [{(\cos x)^{\cos x}}] = \cos x[\log (\cos x)]\]

Substituting the value in equation (1)

\[\log y = \cos x[\log (\cos x)]\]

Now we differentiate on both sides of the equation with respect to x.

\[ \Rightarrow \dfrac{d}{{dx}}\left( {\log y} \right) = \dfrac{d}{{dx}}\left( {\cos x[\log (\cos x)]} \right)\]

We will solve the RHS of the equation first.

We apply the product rule of differentiation on RHS of the equation.

Product rule says that \[\dfrac{d}{{dx}}(mn) = m\dfrac{{dn}}{{dx}} + n\dfrac{{dm}}{{dx}}\].

Substitute the values of \[m = \cos x,n = \log (\cos x)\]

\[

\Rightarrow \dfrac{d}{{dx}}\left( {\cos x[\log (\cos x)]} \right) = \cos x\dfrac{{d[\log (\cos x)]}}{{dx}} + [\log (\cos x)]\dfrac{{d(\cos x)}}{{dx}} \\

\Rightarrow \dfrac{d}{{dx}}\left( {\cos x[\log (\cos x)]} \right) = \cos x\dfrac{{d[\log (\cos x)]}}{{dx}} + [\log (\cos x)]( - \sin x) \\

\]

… (2)

Now we have to apply chain rule for differentiation of the term \[[\log (\cos x)]\].

According to chain rule \[\dfrac{d}{{dx}}\left[ {f(g(x))} \right] = f'(g(x)).g'(x)\]where \[f'\]denotes differentiation of function f with respect to x and \[g'\]denotes differentiation of function g with respect to x.

Here substituting the values of \[f(x) = \log (x),g(x) = \cos x\]

\[

\dfrac{d}{{dx}}\left[ {\log (\cos x)} \right] = \dfrac{{d[\log (\cos x)]}}{{dx}}.\dfrac{{d(\cos x)}}{{dx}} \\

\dfrac{d}{{dx}}\left[ {\log (\cos x)} \right] = \dfrac{1}{{\cos x}}.( - \sin x) \\

\]

Substitute the value in equation (2)

\[ \Rightarrow \dfrac{d}{{dx}}\left( {\cos x[\log (\cos x)]} \right) = \cos x \times \dfrac{1}{{\cos x}} \times ( - \sin x) + [\log (\cos x)]( - \sin x)\]

Cancel out the common terms from numerator and denominator.

\[ \Rightarrow \dfrac{d}{{dx}}\left( {\cos x[\log (\cos x)]} \right) = - \sin x - \sin x[\log (\cos x)]\]

Now we can take \[ - \sin x\] common and write the terms in RHS.

\[ \Rightarrow \dfrac{d}{{dx}}\left( {\cos x[\log (\cos x)]} \right) = - \sin x\{ 1 + \log (\cos x)\} \]

Now solving LHS of the equation we get

\[ \Rightarrow \dfrac{d}{{dx}}\left( {\log y} \right) = \dfrac{1}{y} \times \dfrac{{dy}}{{dx}}\] {Applying chain rule}

Now equating both LHS and RHS of the equation we get

\[ \Rightarrow \dfrac{1}{y} \times \dfrac{{dy}}{{dx}} = - \sin x\{ 1 + \log (\cos x)\} \]

Cross multiplying the value of y to RHS of the equation.

\[

\Rightarrow \dfrac{{dy}}{{dx}} = - \sin x\{ 1 + \log (\cos x)\} \times y \\

\Rightarrow \dfrac{{dy}}{{dx}} = - \sin x\{ 1 + \log (\cos x)\} \times {(\cos x)^{\cos x}} \\

\]

Thus, differentiation of \[{(\cos x)^{\cos x}}\] is \[ - \sin x\{ 1 + \log (\cos x)\} \times {(\cos x)^{\cos x}}\].

**Note:**Students are likely to make mistake in solving this question as they assume the power as normal power and try to differentiate directly through the way \[\dfrac{{d({x^n})}}{{dx}} = n{x^{n - 1}}\] which is wrong.

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