Question

Find derivatives of sin inverse x using chain rule

Answer 1

$\frac{1}{\sqrt{1-x^2}}$

Answer 2

To find the derivative of $\sin^{-1}x$ using the chain rule, we follow these steps:

1. Let $y = \sin^{-1}x$. This implies $x = \sin y$.

2. Differentiate both sides of the equation $x = \sin y$ with respect to $x$.

   • On the left side, $\frac{d}{dx}(x) = 1$.
   • On the right side, we differentiate $\sin y$ with respect to $x$. Since $y$ is a function of $x$, we apply the chain rule: $\frac{d}{dx}(\sin y) = \frac{d}{dy}(\sin y) \cdot \frac{dy}{dx} = \cos y \cdot \frac{dy}{dx}$.

3. Equating the derivatives from both sides: $1 = \cos y \cdot \frac{dy}{dx}$

4. Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{1}{\cos y}$

5. Now, we need to express $\cos y$ in terms of $x$. We know that $\sin y = x$. Using the trigonometric identity $\sin^2 y + \cos^2 y = 1$: $\cos^2 y = 1 - \sin^2 y$

   Substitute $\sin y = x$: $\cos^2 y = 1 - x^2$ $\cos y = \pm \sqrt{1 - x^2}$

   For the principal value branch of $\sin^{-1}x$, the range of $y$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$. In this interval, $\cos y$ is non-negative. Therefore, we take the positive square root: $\cos y = \sqrt{1 - x^2}$

6. Substitute this expression for $\cos y$ back into the derivative: $\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}$

This derivative is valid for $x \in (-1, 1)$.