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# Differentiate ${(\cos x)^{\cos x}}$ with respect to x.

Last updated date: 11th Aug 2024
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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, $\log {(m)^n} = n(\log m)$

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.