Question

Differentiate the function ${{\left( \sin x \right)}^{x}}$ with respect to x.

Answer 1

We need to find the value of $\dfrac{dy}{dx}$ where $y={{\left( \sin x \right)}^{x}}$. To make the equation more differentiable using different methods, we take logarithm on both sides of the equation to get $\log y=x\log \left( \sin x \right)$. Then we take differentiation on both sides. Using the formula of differentiation of $\log x$ and the chain rule we find the differentiation value. We replace the value of y in the final answer to get the solution of the problem.

Complete step-by-step solution
Let’s assume $y={{\left( \sin x \right)}^{x}}$. We need to differentiate y with respect to x. We are trying to find $\dfrac{dy}{dx}$.
First, we change the form of “y” taking logarithms on both sides.
So, $y={{\left( \sin x \right)}^{x}}\Rightarrow \log y=\log {{\left( \sin x \right)}^{x}}$.
We have a logarithmic operation ${{\log }_{a}}{{b}^{n}}=n{{\log }_{a}}b$. We apply that on $\log y=\log {{\left( \sin x \right)}^{x}}$.
$\begin{aligned} & \log y=\log {{\left( \sin x \right)}^{x}} \\\ & \Rightarrow \log y=x\log \left( \sin x \right) \\\ \end{aligned}$.
Now we apply differentiation on the equation $\log y=x\log \left( \sin x \right)$.
$\dfrac{d}{dx}\left( \log y \right)=\dfrac{d}{dx}\left[ x\log \left( \sin x \right) \right]$.
Differentiation of $\log x$ is $\dfrac{1}{x}$. We also apply chain rule to find the right-side differentiation of the equation.
$\begin{aligned} & \dfrac{d}{dx}\left( \log y \right)=\dfrac{d}{dx}\left[ x\log \left( \sin x \right) \right] \\\ & \Rightarrow \dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{x}{\sin x}\times \cos x+\log \left( \sin x \right) \\\ & \Rightarrow \dfrac{1}{y}\dfrac{dy}{dx}=x\cot x+\log \left( \sin x \right) \\\ \end{aligned}$.
We need to find the value of $\dfrac{dy}{dx}$. So, we multiply with y on both sides and get
$\begin{aligned} & \dfrac{1}{y}\dfrac{dy}{dx}=x\cot x+\log \left( \sin x \right) \\\ & \Rightarrow \dfrac{dy}{dx}=y\left[ x\cot x+\log \left( \sin x \right) \right] \\\ \end{aligned}$.
We put the value of y in the equation.
$\begin{aligned} & \dfrac{dy}{dx}=y\left[ x\cot x+\log \left( \sin x \right) \right] \\\ & \Rightarrow \dfrac{dy}{dx}={{\left( \sin x \right)}^{x}}\left[ x\cot x+\log \left( \sin x \right) \right] \\\ \end{aligned}$.
So, the differentiation of the given equation ${{\left( \sin x \right)}^{x}}$ with respect to x is $\dfrac{dy}{dx}={{\left( \sin x \right)}^{x}}\left[ x\cot x+\log \left( \sin x \right) \right]$.

Note: We need to remember that we could not have solved it using the differentiation of $y={{a}^{x}}$ form as in the formula value of a has to be constant. Here the value of a was variable as $a=\sin x$. So, we needed to change the form of the exponential into multiplication. That’s why we used the logarithm to find the simplified form.