# Find the derivative of the given function ${\left( {\cos x} \right)^x}$ .

Answer

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Hint: In order to solve this question we will use a technique called logarithmic differentiation. Also we have to use product rules. Thus, we get our desired answer.

Complete step-by-step answer:

Now given function is,

${\left( {\cos x} \right)^x}$

And we have to find its derivative.

And we will find its derivative by using a technique called logarithmic differentiation.

Logarithmic differentiation- In calculus, logarithmic differentiation or differentiation by taking logarithms is a method used to differentiate functions by employing the logarithmic derivative of function $f$,

$

{\left( {\ln f} \right)^\prime } = \dfrac{{f'}}{f} \\

\Rightarrow f' = f \times {\left( {\ln f} \right)^\prime } \\

$

Now let $y = {\left( {\cos x} \right)^x}$

Now taking log both sides we get,

$\ln \left( y \right) = \ln {\left( {\cos x} \right)^x}$

Or we can write the above equation as,

$\ln \left( y \right) = x\ln \left( {\cos x} \right)$ (By law of logarithms.)

Now differentiate both sides we get,

$\dfrac{{d\left( {\ln \left( y \right)} \right)}}{{dx}} = \dfrac{{d\left( {x\ln \left( {\cos x} \right)} \right)}}{{dx}}$

Now applying the product rule we get,

$\dfrac{{dy}}{{dx}} \times \dfrac{{d\left( {\ln \left( y \right)} \right)}}{{dy}} = \ln \left( {\cos x} \right) \times \dfrac{{dx}}{{dx}} + x\dfrac{{d\left( {\ln \left( {\cos x} \right)} \right)}}{{dx}}$ -----(1)

Now we know that

$\dfrac{{d\left( {\ln \left( y \right)} \right)}}{d} = \dfrac{1}{y}$ ------(2)

Now put (2) in (1) we get,

$\dfrac{{dy}}{{dx}} \times \dfrac{1}{y} = \ln \left( {\cos x} \right) + x\dfrac{{d\left( {\ln \left( {\cos x} \right)} \right)}}{{dx}}$ ----(3) $\left( {\because \dfrac{{dx}}{{dx}} = 1} \right)$

Also,

$

\dfrac{{d\left( {\ln \left( {\cos x} \right)} \right)}}{{dx}} = \dfrac{{d\left( {\ln \left( {\cos x} \right)} \right)}}{{d\left( {\cos x} \right)}} \times \dfrac{{d\left( {\cos x} \right)}}{{dx}} \\

= \dfrac{1}{{\cos x}} \times - \sin x \\

$

$ = \dfrac{{ - \sin x}}{{\cos x}}$ -----(4)

Putting the value of (4) in (3) we get,

$\dfrac{{dy}}{{dx}} \times \dfrac{1}{y} = \ln \left( {\cos x} \right) - \dfrac{{x\sin x}}{{\cos x}}$

Or $\dfrac{{dy}}{{dx}} = y\left( {\ln \left( {\cos x} \right) - \dfrac{{x\sin x}}{{\cos x}}} \right)$

Or $\dfrac{{dy}}{{dx}} = {\left( {\cos x} \right)^x}\left( {\ln \left( {\cos x} \right) - \dfrac{{x\sin x}}{{\cos x}}} \right)$

Thus, the required answer is ${\left( {\cos x} \right)^x}\left( {\ln \left( {\cos x} \right) - \dfrac{{x\sin x}}{{\cos x}}} \right)$ .

Note: Whenever we face such types of questions the key concept is that take logarithm function both sides and then solve the equation formed. Like, in this question we simply assume a variable which is equal to the given function then we take logarithmic function both sides and then we solve the equation formed and thus we get our answer.

Complete step-by-step answer:

Now given function is,

${\left( {\cos x} \right)^x}$

And we have to find its derivative.

And we will find its derivative by using a technique called logarithmic differentiation.

Logarithmic differentiation- In calculus, logarithmic differentiation or differentiation by taking logarithms is a method used to differentiate functions by employing the logarithmic derivative of function $f$,

$

{\left( {\ln f} \right)^\prime } = \dfrac{{f'}}{f} \\

\Rightarrow f' = f \times {\left( {\ln f} \right)^\prime } \\

$

Now let $y = {\left( {\cos x} \right)^x}$

Now taking log both sides we get,

$\ln \left( y \right) = \ln {\left( {\cos x} \right)^x}$

Or we can write the above equation as,

$\ln \left( y \right) = x\ln \left( {\cos x} \right)$ (By law of logarithms.)

Now differentiate both sides we get,

$\dfrac{{d\left( {\ln \left( y \right)} \right)}}{{dx}} = \dfrac{{d\left( {x\ln \left( {\cos x} \right)} \right)}}{{dx}}$

Now applying the product rule we get,

$\dfrac{{dy}}{{dx}} \times \dfrac{{d\left( {\ln \left( y \right)} \right)}}{{dy}} = \ln \left( {\cos x} \right) \times \dfrac{{dx}}{{dx}} + x\dfrac{{d\left( {\ln \left( {\cos x} \right)} \right)}}{{dx}}$ -----(1)

Now we know that

$\dfrac{{d\left( {\ln \left( y \right)} \right)}}{d} = \dfrac{1}{y}$ ------(2)

Now put (2) in (1) we get,

$\dfrac{{dy}}{{dx}} \times \dfrac{1}{y} = \ln \left( {\cos x} \right) + x\dfrac{{d\left( {\ln \left( {\cos x} \right)} \right)}}{{dx}}$ ----(3) $\left( {\because \dfrac{{dx}}{{dx}} = 1} \right)$

Also,

$

\dfrac{{d\left( {\ln \left( {\cos x} \right)} \right)}}{{dx}} = \dfrac{{d\left( {\ln \left( {\cos x} \right)} \right)}}{{d\left( {\cos x} \right)}} \times \dfrac{{d\left( {\cos x} \right)}}{{dx}} \\

= \dfrac{1}{{\cos x}} \times - \sin x \\

$

$ = \dfrac{{ - \sin x}}{{\cos x}}$ -----(4)

Putting the value of (4) in (3) we get,

$\dfrac{{dy}}{{dx}} \times \dfrac{1}{y} = \ln \left( {\cos x} \right) - \dfrac{{x\sin x}}{{\cos x}}$

Or $\dfrac{{dy}}{{dx}} = y\left( {\ln \left( {\cos x} \right) - \dfrac{{x\sin x}}{{\cos x}}} \right)$

Or $\dfrac{{dy}}{{dx}} = {\left( {\cos x} \right)^x}\left( {\ln \left( {\cos x} \right) - \dfrac{{x\sin x}}{{\cos x}}} \right)$

Thus, the required answer is ${\left( {\cos x} \right)^x}\left( {\ln \left( {\cos x} \right) - \dfrac{{x\sin x}}{{\cos x}}} \right)$ .

Note: Whenever we face such types of questions the key concept is that take logarithm function both sides and then solve the equation formed. Like, in this question we simply assume a variable which is equal to the given function then we take logarithmic function both sides and then we solve the equation formed and thus we get our answer.

Last updated date: 27th May 2023

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