The derivative of x^tanx (x to the power tanx) is denoted by d/dx(x^{tanx}) and its value is equal to x^{tanx}[tanx/x + sec^{2}x log_{e}x].

The derivative formula of x^{tanx} is given by

d/dx(x^{tanx}) = x^{tanx}[tanx/x + sec^{2}x log_{e}x].

Let us now learn how to differentiate x^{tanx}.

## Differentiate x^{tanx}

**Question:** Prove that d/dx(x^{tanx}) = x^{tanx}[tanx/x + sec^{2}x logx].

**Answer:**

Let us put

y=x^{tanx}.

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

log_{e} y = log_{e} x^{tanx}

⇒ log_{e}y = tanx log_{e}x, as we know the logarithm formula log_{a}b^{n} = n log_{a}b.

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(log_{e}x) =1/x and d/dx(tan x)= sec^{2}x.

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

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

So the derivative of x^{tanx} (x to the power tanx) is equal to x^{tanx}[tanx/x + sec^{2}x logx], and it is obtained by the logarithmic differentiation.

**More Derivatives:**

Derivative of x^{x} | Derivative of x^{sinx}

Derivative of sinx/x | Derivative of x^{logx}

## FAQs

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

**Answer:** The derivative of x^tanx (x raised to the power tanx) is equal to x^{tanx}[tanx/x + sec^{2}x logx].