Q.Find the derivative of $f$ given by $f(x) = \sin^{-1} x$ assuming it exists.

The key idea is to use implicit differentiation on the inverse relation $x = \sin y$, then apply the Pythagorean identity to express $\cos y$ in terms of $x$.

1. Let $y = \sin^{-1} x$, so $x = \sin y$, where $y \in [-\pi/2, \pi/2]$.
2. Differentiate both sides with respect to $x$: $1 = \cos y \cdot \frac{dy}{dx}$. …

The key insight here is that $\sin^{-1} x$ (also written as $\arcsin x$) is the inverse of the sine function restricted to $[-\pi/2, \pi/2]$. To differentiate an inverse function, we don't need a separate formula — we can use implicit differentiation on the original relationship.

1. Set up the inverse relationship.

   Let $y = \sin^{-1} x$. By definition, this means $x = \sin y$, where $y$ lies in $[-\pi/2, \pi/2]$. This domain restriction is crucial because it tells us the sign of $\cos y$ later.

2. Differentiate both sides with respect to $x$.

   Since $y$ is a function of $x$, we differentiate implicitly:

3. Solve for $\frac{dy}{dx}$. Provided $\cos y \neq 0$, we get:

4. Express $\cos y$ in terms of $x$. We know $x = \sin y$. Using the Pythagorean identity $\sin^2 y + \cos^2 y = 1$, we have:

Taking the square root gives $\cos y = \pm \sqrt{1 - x^2}$. …

Method: Deriving the Derivative of an Inverse Trigonometric Function

Use this method to find (or re-derive) the derivative of any inverse trig function such as $\sin^{-1}x$, $\cos^{-1}x$, or $\tan^{-1}x$, by converting the inverse relation back into the original trig function and differentiating implicitly.

Steps

Step 1: Rewrite the inverse relation as a direct trig equation

If $y = \sin^{-1}x$, this means, by definition, $x = \sin y$, with $y$ restricted to the function's principal value range (here $\left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$).

Step 2: Differentiate both sides implicitly with respect to $x$

Step 3: Solve for $\dfrac{dy}{dx}$ in terms of $y$

Step 4: Convert back to a function of $x$ using a Pythagorean identity

Since $x = \sin y$, use $\cos^2 y = 1 - \sin^2 y = 1 - x^2$, so $\cos y = \pm\sqrt{1-x^2}$.

Step 5: Fix the sign using the principal-range restriction …

Common Mistakes

Mistake 1: Taking the wrong sign for $\cos y$

Why it's wrong: $\cos^2 y = 1 - x^2$ has two square roots, $\pm\sqrt{1-x^2}$, and picking the negative one gives the wrong final derivative. The correct sign depends entirely on the range that $y = \sin^{-1}x$ is restricted to. Correct approach: since the principal value range of $\sin^{-1}x$ is $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$, where cosine is never negative, always take the positive root, $\cos y = \sqrt{1-x^2}$.

Mistake 2: Forgetting the domain restriction on the final answer …