# Derivative of x^tanx (x to the power tanx)

The derivative of x^tanx (x to the power tanx) is denoted by d/dx(xtanx) and its value is equal to xtanx[tanx/x + sec2x logex].

The derivative formula of xtanx is given by

d/dx(xtanx) = xtanx[tanx/x + sec2x logex].

Let us now learn how to differentiate xtanx.

Table of Contents

## Differentiate xtanx

Question: Prove that d/dx(xtanx) = xtanx[tanx/x + sec2x logx].

Answer:

Let us put

y=xtanx.

Here we need to find dy/dx. Taking logarithms on both sides, we get that

loge y = loge xtanx

⇒ logey = tanx logex, as we know the logarithm formula logabn = n logab.

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

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

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

⇒ $\dfrac{1}{y} \dfrac{dy}{dx}$ $=\tan x \cdot \dfrac{1}{x}+\log_e x \sec^2 x$ as we know d/dx(logex) =1/x and d/dx(tan x)= sec2x.

⇒ $\dfrac{dy}{dx}=y(\dfrac{\tan x}{x}+\sec^2 x\log_e x)$

⇒ $\dfrac{dy}{dx}=x^{\tan x}(\dfrac{\tan x}{x}+\sec^2 x \log_e x)$ as y=xtanx.

So the derivative of xtanx (x to the power tanx) is equal to xtanx[tanx/x + sec2x logx], and it is obtained by the logarithmic differentiation.

More Derivatives:

Derivative of xcosx

Derivative of x3/2

## FAQs

### Q1: What is the derivative of x^tanx?

Answer: The derivative of x^tanx (x raised to the power tanx) is equal to xtanx[tanx/x + sec2x logx].

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