# 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.

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