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Question: How do you differentiate \(y = {\left( {\cos 7x} \right)^x}\)?...

How do you differentiate y=(cos7x)xy = {\left( {\cos 7x} \right)^x}?

Explanation

Solution

First of all we will take logarithmic function on both the sides and then we will differentiate the function. Then, we will modify so that we have only dy/dx on the left and rest on right.

Complete step by step solution:
We are given that we are required to find the differentiation of y=(cos7x)xy = {\left( {\cos 7x} \right)^x}.
Taking logarithmic function on both the sides of above equation, we will then obtain the following equation with us:-
logy=log(cos7x)x\Rightarrow \log y = \log {\left( {\cos 7x} \right)^x}
Simplifying the above equation, we get the following equation with us:-
logy=xlog(cos7x)\Rightarrow \log y = x\log \left( {\cos 7x} \right)
Differentiating both the sides of above equation with respect to x, we will obtain the following equation with us:-
\Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left\\{ {x\log \left( {\cos 7x} \right)} \right\\}
Now, we will use the chain rule for differentiation and then get the following equation with us:-
1ydydx=log(cos7x)+xcos7x(7sin7x)\Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = \log \left( {\cos 7x} \right) + \dfrac{x}{{\cos 7x}}\left( {7\sin 7x} \right)
Taking y from division in the left hand side to multiplication in the right hand side, we will obtain the following equation with us:-
dydx=ylog(cos7x)+7xysin7xcos7x\Rightarrow \dfrac{{dy}}{{dx}} = y\log \left( {\cos 7x} \right) + \dfrac{{7xy\sin 7x}}{{\cos 7x}}
Putting the value of y from the given equation, we will then obtain the following equation with us:-
dydx=(cos7x)xlog(cos7x)+7x(cos7x)xsin7xcos7x\Rightarrow \dfrac{{dy}}{{dx}} = {\left( {\cos 7x} \right)^x}\log \left( {\cos 7x} \right) + 7x{\left( {\cos 7x} \right)^x}\dfrac{{\sin 7x}}{{\cos 7x}}
The equation mentioned above can be written in the form of the following equation as well:-
dydx=(cos7x)xlog(cos7x)+7xtan7x(cos7x)x\Rightarrow \dfrac{{dy}}{{dx}} = {\left( {\cos 7x} \right)^x}\log \left( {\cos 7x} \right) + 7x\tan 7x{\left( {\cos 7x} \right)^x}
Re – writing the above equation by arranging its terms, we will then obtain the following equation with us:-

\Rightarrow \dfrac{{dy}}{{dx}} = {\left( {\cos 7x} \right)^x}\left\\{ {\log \left( {\cos 7x} \right) + 7x\tan 7x} \right\\}

Thus, we have the required answer.

Note:
The students must note the following facts and commit them to the memory which were used in the solution given above:-

  1. ddx(logx)=1x\dfrac{d}{{dx}}\left( {\log x} \right) = \dfrac{1}{x}
  2. The differentiation of logarithmic function gives us the inverse function.
  3. If we are given two functions u and v in the form u.v, then its differentiation is given by the following expression: (uv)=uv+uv\left( {uv} \right)' = u'v + uv'
  4. This is known as the chain rule of differentiation as we mentioned in the solution given above.
  5. ddx(sinx)=cosx\dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x
  6. This implies that the differentiation of sine of any angle gives us the cosine of the same angle.