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Question: Derivative of \({(x\cos x)^x}\) with respect to \(x\) is A.\({(x\cos x)^x}\) \([(\log x + 1) - \\...

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

Explanation

Solution

- 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)y = f(x) = {[u(x)]^{v(x)}} and take the logarithm of both sides to get log⁡(y)=v(x)log⁡u(x)\log (y) = v(x)\log u(x). Further there are detailed explanations of each step.

Complete step-by-step answer:
Step1 Let y=(xcos⁡x)xy = {(x\cos x)^x}
Take the logarithm of both sides
log⁡y=log⁡(xcos⁡x)x\log y = \log {(x\cos x)^x}
Using property of log;
log⁡y=xlog⁡(xcos⁡x)\log y = x\log (x\cos x)
Differentiating both sides with respect to x, we get
1ydydx=xddxlog⁡(xcos⁡x)+log⁡(xcos⁡x)ddx(x)\dfrac{1}{y}\dfrac{{dy}}{{dx}} = x\dfrac{d}{{dx}}\\{ \log (x\cos x)\\} + \log (x\cos x)\dfrac{d}{{dx}}(x)
Step 2
⇒1ydydx=x×1xcos⁡xx(−sin⁡x)+cos⁡x.1+log⁡(xcos⁡x).1 ⇒1ydydx=1cos⁡xx(−sin⁡x)+cos⁡x+log⁡(xcos⁡x) ⇒1ydydx=x(−sin⁡x)cos⁡x+cos⁡xcos⁡x+log⁡(xcos⁡x)  \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 1ydydx=x(−sin⁡x)cos⁡x+1+log⁡x+log⁡cos⁡x\dfrac{1}{y}\dfrac{{dy}}{{dx}} = \dfrac{{x( - \sin x)}}{{\cos x}} + 1 + \log x + \log \cos x
⇒\Rightarrow 1ydydx=(log⁡x+1)+log⁡cos⁡x+xcos⁡x(−sin⁡x)\dfrac{1}{y}\dfrac{{dy}}{{dx}} = (\log x + 1) + \\{ \log \cos x + \dfrac{x}{{\cos x}}( - \sin x)\\}
⇒dydx=y[(log⁡x+1)+log⁡cos⁡x+xcos⁡x(−sin⁡x) ⇒dydx=(xcos⁡x)x[(log⁡x+1)+log⁡cos⁡x+xcos⁡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 dydx=dydududx\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}}\dfrac{{du}}{{dx}}