Question

Derivative of ${(x\cos x)^x}$ with respect to $x$ is
A.${(x\cos x)^x}$$[(\log x + 1) - \{ \log \cos x + \dfrac{x}{{\cos x}}(\sin x)\} ]$
B.${(x\cos x)^x}$$[(\log x + 1) + \{ \log \cos x + \dfrac{x}{{\cos x}}( - \sin x)\} ]$
C.${(x\cos x)^x}$$[(\log x + 1) + \{ \log \sin x + \dfrac{x}{{\cos x}}(\cos x)\} ]$
D.None of these

Answer 1

Hint: - The derivative of a function of a real variable measures the sensitivity to change of the function value with respect to a change in its argument. Derivatives are a fundamental tool of calculus.
To solve this problem, we have to differentiate functions of the form $y = f(x) = {[u(x)]^{v(x)}}$ and take the logarithm of both sides to get $\log (y) = v(x)\log u(x)$. Further there are detailed explanations of each step.

Complete step-by-step answer:
Step1 Let $y = {(x\cos x)^x}$
Take the logarithm of both sides
$\log y = \log {(x\cos x)^x}$
Using property of log;
$\log y = x\log (x\cos x)$
Differentiating both sides with respect to x, we get
$\dfrac{1}{y}\dfrac{{dy}}{{dx}} = x\dfrac{d}{{dx}}\{ \log (x\cos x)\} + \log (x\cos x)\dfrac{d}{{dx}}(x)$
Step 2
$
\Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = x \times \dfrac{1}{{x\cos x}}\{ x( - \sin x) + \cos x.1\} + \log (x\cos x).1 \\
\Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = \dfrac{1}{{\cos x}}\{ x( - \sin x) + \cos x\} + \log (x\cos x) \\
\Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = \dfrac{{x( - \sin x)}}{{\cos x}} + \dfrac{{\cos x}}{{\cos x}} + \log (x\cos x) \\
$
Cancelling cosx with cosx, we get,
$\Rightarrow$ $\dfrac{1}{y}\dfrac{{dy}}{{dx}} = \dfrac{{x( - \sin x)}}{{\cos x}} + 1 + \log x + \log \cos x$
$\Rightarrow$ $\dfrac{1}{y}\dfrac{{dy}}{{dx}} = (\log x + 1) + \{ \log \cos x + \dfrac{x}{{\cos x}}( - \sin x)\} $
$
\Rightarrow \dfrac{{dy}}{{dx}} = y[(\log x + 1) + \{ \log \cos x + \dfrac{x}{{\cos x}}( - \sin x)\} \\
\Rightarrow \dfrac{{dy}}{{dx}} = {(x\cos x)^x}[(\log x + 1) + \{ \log \cos x + \dfrac{x}{{\cos x}}( - \sin x)\} \\
$
So, B is the correct answer.

Note: Logarithmic differentiation is also useful when we are to differentiate a product or quotient of more than two functions.
Chain rule – It is the process that helps us to find the derivative of a composite function.
If y is a differentiable function of u and u is a differentiable function then $\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}}\dfrac{{du}}{{dx}}$