The derivative of x^{x} is equal to x^{sinx}[sinx/x + cosx logx]. In this post, we will learn how to differentiate x^sinx (x to the power sinx).

## Derivative of x^{sinx} Formula

x^{sinx }Derivative Formula: The formula of x^{sinx} derivative is given by

d/dx(x^{sinx}) = x^{sinx}[sinx/x + cosx logx]

Let us now give a proof of this fact.

## Derivative of x^{sinx} Proof

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

*Answer:*

Let y=x^{sinx}. We need to find dy/dx.

Taking the natural logarithms both sides, we have

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

⇒ log_{e}y =sinx log_{e}x.

Taking d/dx on both sides, we have

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

Now, applying the product rule of derivatives, it follows that

$\dfrac{1}{y} \dfrac{dy}{dx}$ $=\sin x\dfrac{d}{dx}{\log_e x}+\log_e x\dfrac{d}{dx}(\sin x)$

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

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

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

Thus, the derivative of x^{sinx} (x to the sinx) is equal to x^{sinx}[sinx/x + cosx logx]. This is proved by the logarithmic differentiation.

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## FAQs

**Q1: What is the derivative of x^sinx?**

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