Question

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

Answer 1

Hint: To obtain the derivative of a given function we will use implicit differentiation. Firstly start by taking the sine function on the left side of the equation then differentiate both sides by $x$. Then we will simplify it further by using the original function and some trigonometric identities and get our desired answer.

Complete step-by-step answer:
We have to find the derivative of
$y=\arcsin \left( \sqrt{x} \right)$
Start by writing the simplified form of right side value as below:
$y={{\sin }^{-1}}\left( \sqrt{x} \right)$
Now take the sine term on the left hand side as below:
$\sin \left( y \right)=\sqrt{x}$……$\left( 1 \right)$
Next, using implicit differentiation on above function we get,
$\dfrac{d}{dx}\left( \sin \left( y \right) \right)=\dfrac{d}{dx}\left( \sqrt{x} \right)$
The differentiation of sine is cosine and $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$ so we get,
$\begin{align}
  & \Rightarrow \cos \left( y \right)\dfrac{dy}{dx}=\dfrac{1}{2}{{x}^{\dfrac{1}{2}-1}} \\
 & \Rightarrow \cos \left( y \right)\dfrac{dy}{dx}=\dfrac{1}{2}{{x}^{-\dfrac{1}{2}}} \\
\end{align}$
$\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{2\sqrt{x}\times \cos \left( y \right)}$……..$\left( 2 \right)$
Now we have to find the value of $\cos \left( y \right)$ in term of $x$ so we will use the below identity,
${{\sin }^{2}}y+{{\cos }^{2}}y=1$
Substitute the value from equation (1) in above we get,
$\begin{align}
  & \Rightarrow {{\left( \sqrt{x} \right)}^{2}}+{{\cos }^{2}}y=1 \\
 & \Rightarrow {{\cos }^{2}}y=1-x \\
 & \therefore \cos y=\pm \sqrt{1-x} \\
\end{align}$
We will take only the positive square root value for $\cos y$ as ${{\sin }^{-1}}\sqrt{x}$ have output between $\dfrac{-\pi }{2}$ and $\dfrac{\pi }{2}$ and we know that cosine function is positive in this interval.
Substitute the positive value from above value in equation (2) as follows:
$\begin{align}
  & \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{2\sqrt{x}\times \sqrt{1-x}} \\
 & \therefore \dfrac{dy}{dx}=\dfrac{1}{2\sqrt{x\left( 1-x \right)}} \\
\end{align}$

Hence derivative of $y=\arcsin \left( \sqrt{x} \right)$ is $\dfrac{dy}{dx}=\dfrac{1}{2\sqrt{x\left( 1-x \right)}}$

Note: Differentiation is a process of finding derivatives of a function. If we have two unknown variables then the rate of change of one with respect to the other is known as the derivative. Differentiation of trigonometric function is different from the other differentiation as it is a very vast topic with many identities and exceptions in it. Implicit differentiation is done when differentiating one term by another doesn't lead to the answer directly.