
How do you differentiate $y = {\left( {\cos 7x} \right)^x}$?
Answer
535.5k+ views
Hint: 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 $y = {\left( {\cos 7x} \right)^x}$.
Taking logarithmic function on both the sides of above equation, we will then obtain the following equation with us:-
$ \Rightarrow \log y = \log {\left( {\cos 7x} \right)^x}$
Simplifying the above equation, we get the following equation with us:-
$ \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:-
\[ \Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = \log \left( {\cos 7x} \right) + \dfrac{x}{{\cos 7x}}\left( {7\sin 7x} \right)\]
Taking y from division in the left hand side to multiplication in the right hand side, we will obtain the following equation with us:-
\[ \Rightarrow \dfrac{{dy}}{{dx}} = y\log \left( {\cos 7x} \right) + \dfrac{{7xy\sin 7x}}{{\cos 7x}}\]
Putting the value of y from the given equation, we will then obtain the following equation with us:-
\[ \Rightarrow \dfrac{{dy}}{{dx}} = {\left( {\cos 7x} \right)^x}\log \left( {\cos 7x} \right) + 7x{\left( {\cos 7x} \right)^x}\dfrac{{\sin 7x}}{{\cos 7x}}\]
The equation mentioned above can be written in the form of the following equation as well:-
\[ \Rightarrow \dfrac{{dy}}{{dx}} = {\left( {\cos 7x} \right)^x}\log \left( {\cos 7x} \right) + 7x\tan 7x{\left( {\cos 7x} \right)^x}\]
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) $\dfrac{d}{{dx}}\left( {\log x} \right) = \dfrac{1}{x}$
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: $\left( {uv} \right)' = u'v + uv'$
4) This is known as the chain rule of differentiation as we mentioned in the solution given above.
5) $\dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x$
6) This implies that the differentiation of sine of any angle gives us the cosine of the same angle.
Complete step by step solution:
We are given that we are required to find the differentiation of $y = {\left( {\cos 7x} \right)^x}$.
Taking logarithmic function on both the sides of above equation, we will then obtain the following equation with us:-
$ \Rightarrow \log y = \log {\left( {\cos 7x} \right)^x}$
Simplifying the above equation, we get the following equation with us:-
$ \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:-
\[ \Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = \log \left( {\cos 7x} \right) + \dfrac{x}{{\cos 7x}}\left( {7\sin 7x} \right)\]
Taking y from division in the left hand side to multiplication in the right hand side, we will obtain the following equation with us:-
\[ \Rightarrow \dfrac{{dy}}{{dx}} = y\log \left( {\cos 7x} \right) + \dfrac{{7xy\sin 7x}}{{\cos 7x}}\]
Putting the value of y from the given equation, we will then obtain the following equation with us:-
\[ \Rightarrow \dfrac{{dy}}{{dx}} = {\left( {\cos 7x} \right)^x}\log \left( {\cos 7x} \right) + 7x{\left( {\cos 7x} \right)^x}\dfrac{{\sin 7x}}{{\cos 7x}}\]
The equation mentioned above can be written in the form of the following equation as well:-
\[ \Rightarrow \dfrac{{dy}}{{dx}} = {\left( {\cos 7x} \right)^x}\log \left( {\cos 7x} \right) + 7x\tan 7x{\left( {\cos 7x} \right)^x}\]
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) $\dfrac{d}{{dx}}\left( {\log x} \right) = \dfrac{1}{x}$
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: $\left( {uv} \right)' = u'v + uv'$
4) This is known as the chain rule of differentiation as we mentioned in the solution given above.
5) $\dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x$
6) This implies that the differentiation of sine of any angle gives us the cosine of the same angle.
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