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

The derivative of x^{cosx} is denoted by d/dx(x^{cosx}) and its formula is given below:

d/dx(x^{cosx}) = x^{cosx}[cosx/x – sinx log_{e}x].

Let us now give a proof of x^{cosx} derivative.

## Differentiate x^{cosx}

**Question:** Prove that d/dx(x^{cosx}) = x^{cosx}[cosx/x – sinx logx].

**Answer:**

To differentiate x^{cosx}, we will use the logarithmic differentiation. Let us put

y=x^{cosx}.

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

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

⇒ log_{e}y = cosx log_{e}x, using the logarithm formula: log_{a}b^{n} = n log_{a}b.

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

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

⇒ $\dfrac{1}{y} \dfrac{dy}{dx}$ $=\cos x\dfrac{d}{dx}(\log_e x)+\log_e x\dfrac{d}{dx}(\cos x)$, 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 log_{e}x 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=x^{cosx}.

So the derivative of x^{cosx} (x to the cosx) is equal to x^{cosx}[cosx/x – sinx 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^cosx?**

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