Derivative of x^cosx | x^cosx Derivative

The derivative of x^cosx (x to the power cosx) is equal to xcosx[cosx/x – sinx logex]. Here we learn how to differentiate x^cosx.

The derivative of xcosx is denoted by d/dx(xcosx) and its formula is given below:

d/dx(xcosx) = xcosx[cosx/x – sinx logex].

Let us now give a proof of xcosx derivative.

Table of Contents

Differentiate xcosx

Question: Prove that d/dx(xcosx) = xcosx[cosx/x – sinx logx].

Answer:

To differentiate xcosx, we will use the logarithmic differentiation. Let us put

y=xcosx.

So we need to find dy/dx. Taking the natural logarithms both sides, we get that

loge y = loge xcosx

⇒ logey = cosx logex, using the logarithm formula: logabn = n logab.

Differentiating both sides w.r.t x, we have

$\dfrac{d}{dx}(\log_e y)=\dfrac{d}{dx}(\cos x \log_e x)$

⇒ $\dfrac{1}{y} \dfrac{dy}{dx}$ $=\cos x\dfrac{d}{dx}(\log_e x)+\log_e x\dfrac{d}{dx}(\cos x)$, by the product rule of derivatives.

⇒ $\dfrac{1}{y} \dfrac{dy}{dx}$ $=\cos x \cdot \dfrac{1}{x}+\log_e x (-\sin x)$ as the derivative of logex is 1/x and d/dx(cos x)= -sinx.

⇒ $\dfrac{dy}{dx}=y(\dfrac{\cos x}{x}-\sin x\log_e x)$

⇒ $\dfrac{dy}{dx}=x^{\cos x}(\dfrac{\cos x}{x}-\sin x \log_e x)$ as y=xcosx.

So the derivative of xcosx (x to the cosx) is equal to xcosx[cosx/x – sinx logx], and it is obtained by the logarithmic differentiation.

More Derivatives:

Derivative of x sinx

Derivative of x3/2

FAQs

Q1: What is the derivative of x^cosx?

Answer: The derivative of x^cosx is equal to xcosx[cosx/x – sinx logx].

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