# What is the Derivative of (sinx)^logx

In this blog post, we will find the derivative of sinxlogx, 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 sinxlogx

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 $(\sin x)^{\log x}[ \log x \cot x + \frac{\log \sin x}{x}]$

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## 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].