In this blog post, we will find the derivative of sinx^{logx}, that is, sinx to the power logx, by the product rule of derivatives along with logarithmic differentiation. Let us recall the product rule of derivatives: The derivative of the product function f(x)g(x) is given as follows:

$\dfrac{d}{dx}(f(x)g(x))$ $=f(x) g'(x) + f'(x) g(x)$ $\cdots (\star)$.

Here $’$ denotes the derivative with respect to $x$.

## What is the Derivative of sinx^{logx}

To find the derivative of $(\sin x)^{\log x}$, we have to go through the below steps:

**Step 1:** Let us put $y=(\sin x)^{\log x}$ $\cdots (I)$

So we need to find $\frac{dy}{dx}$.

**Step 2:** We will use logarithmic differentiation. Taking natural logarithms $\ln = \log_e$ of both sides of (I). By doing so we obtain that

$\log_e y = \log_e (\sin x)^{\log_e x}$

$\Rightarrow \log_e y = \log_e x \log_e \sin x$ by the logarithm formula $\log a^k=k\log a$.

**Step 3:** Now, we will apply the above product rule of derivatives $(\star)$on the right side of the above equation. Thus, we have

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

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

$\Rightarrow \frac{1}{y} \frac{dy}{dx}$ $= \log_e x \frac{1}{\sin x} \frac{d}{dx}(\sin x) +\frac{1}{x} \log_e \sin x$ by the chain rule of derivatives.

$\Rightarrow \frac{dy}{dx}$ $=y[\frac{ \log_e x}{\sin x} cos x + \frac{\log_e \sin x}{x}]$

$\Rightarrow \frac{dy}{dx}$ $=(\sin x)^{\log x}[ \log x \cot x + \frac{\log \sin x}{x}]$ as we have $y=(\sin x)^{\log x}$.

**Conclusion:** Thus, the derivative of $(\sin x)^{\log x}$ is equal to $(\sin x)^{\log x}[ \log x \cot x + \frac{\log \sin x}{x}]$

**Also Read:**

**Derivative of 2 ^{x} by First Principle**

**Derivative of x ^{3/2} by First Principle**

## FAQs

**Q1: Find the derivative of (sinx) ^{logx}?**

**Answer:** The derivative of (sinx)^{logx} is equal to (sin x)^{logx}[ logx cotx + log(sinx)/x].