Question

What is the derivative of $y={{\left( \sin x \right)}^{x}}$?

Answer 1

We can notice that the given function is a mix of two functions, i.e. sin x and a function with power of x. So, we can make use of the chain rule. To simplify it better, we will first take log on both sides of the given function and apply property $\log {{a}^{b}}=b\log a$ . Then, we will find the derivative by making use of product rule and chain rule.

Complete step by step solution:
Here we are asked to find out the derivative of $y={{\left( \sin x \right)}^{x}}$. In order to find that, we need to check whether there is any constant term given or not. If there is a constant term then we are not supposed to differentiate the constant term and it should be left as it is.
Here is the list of derivatives of the common trigonometric ratios:

& \left( \sin x \right)^\prime =\cos x \\\ & \left( \cos x \right)^\prime =-\sin x \\\ & \left( \tan x \right)^\prime =\dfrac{1}{{{\cos }^{2}}x} \\\ & \left( \cot x \right)^\prime =-\dfrac{1}{{{\sin }^{2}}x} \\\ & \left( \sec \right)^\prime =\tan x\sec x \\\ & \left( \csc x \right)^\prime =-\cot x\csc x \\\ \end{aligned}$$ Now let us solve the given problem. $$y={{\left( \sin x \right)}^{x}}$$ We will be solving this function using logarithmic differentiation. Taking log on both sides , we get $$\dfrac{dy}{dx}=(\log (\sin x)+x\cot x){{(\sin x)}^{x}}$$ Using the logarithmic differentiation, we get $$y={{\left( \sin x \right)}^{x}}$$ $$\log y=\log {{\left( \sin x \right)}^{x}}$$ Here, we can apply the logarithmic property which is $$\log {{a}^{b}}=b\log a$$. Relating it to our equation, we will be having $$b$$as $$x$$ and $$a$$ as $$\sin x$$. After writing it accordingly, we get $$\Rightarrow x\log \left( \sin x \right)$$ Now let us differentiating using product rule and chain rule: $$\dfrac{1}{y}\dfrac{dy}{dx}=1\log (\sin x)+x[\dfrac{\cos x}{\sin x}]$$ On simplifying $$x\left(\dfrac{\cos x}{\sin x}\right)$$ which is $$\cot x$$ We get, $$\dfrac{1}{y}\dfrac{dy}{dx}=\log (\sin x)+x\cot x$$ Now solve for $$\dfrac{dy}{dx}$$ by multiplying $$y={{\left( \sin x \right)}^{x}}$$ **Then we get the following result- $$\dfrac{dy}{dx}=(\log (\sin x)+x\cot x)(\sin x)x$$** **Note:** The derivation is to be done carefully by following the accurate method such that there would not rise up any complication in finding the derivative. Different rules such as chain rule, product rule, quotient rule etc can be chosen to find out the derivatives.