Question

Differentiate \[{(\cos x)^{\cos x}}\] with respect to x.

Answer 1

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.