Q.If $y = \cos^{-1} x$, Find $\frac{d^2 y}{dx^2}$ in terms of $y$ alone.

Idea: $y=\cos^{-1}x$ means $x=\cos y$. Get $\frac{dy}{dx}$ in terms of $y$, then differentiate again.

First derivative: from $x=\cos y$, $\dfrac{dx}{dy}=-\sin y$, so

Second derivative: differentiate w.r.t. $x$ via $\dfrac{d}{dx}=\dfrac{dy}{dx}\cdot\dfrac{d}{dy}$: …

We want the second derivative of $y=\cos^{-1}x$ expressed in $y$ alone, so we must remove $x$ from the final answer. The cleanest route is to start from the relation $x=\cos y$ and keep working in $y$.

Step 1 — first derivative

From $x = \cos y$, differentiate implicitly with respect to $x$:

Step 2 — second derivative

We differentiate $\frac{dy}{dx}=-\frac{1}{\sin y}$ with respect to $x$. Since it is a function of $y$, use the chain rule $\frac{d}{dx}=\frac{d}{dy}\cdot\frac{dy}{dx}$:

Multiply by $\frac{dy}{dx}=-\frac{1}{\sin y}$:

Step 3 — compact form …

Method: Second Derivative of an Inverse Function, Expressed in Terms of Itself

When asked for $y''$ "in terms of $y$ alone" (not $x$), work from the inverse relation rather than the direct formula for $\frac{d}{dx}\cos^{-1}x$.

Steps

Step 1: Rewrite the inverse relation as $x$ in terms of $y$

For $y=\cos^{-1}x$, this means $x=\cos y$.

Step 2: Differentiate implicitly to get $\dfrac{dy}{dx}$ in terms of $y$

From $x=\cos y$: $1=-\sin y\cdot\frac{dy}{dx}$, so $\frac{dy}{dx}=-\frac{1}{\sin y}$.

Step 3: Differentiate again, routing through $y$ via the chain rule $\dfrac{d}{dx}=\dfrac{dy}{dx}\cdot\dfrac{d}{dy}$ …

Common Mistakes

Mistake 1: Trying to differentiate the standard formula $\dfrac{d}{dx}\cos^{-1}x=-\dfrac{1}{\sqrt{1-x^2}}$ directly.

Why it's wrong: that route produces the answer in terms of $x$, not $y$ — extra work is then needed to convert back, and it's easy to substitute $x=\cos y$ incorrectly partway through. Correct approach: work entirely in $y$ from the start via $x=\cos y$.

Mistake 2: Forgetting to multiply by $\dfrac{dy}{dx}$ a second time when applying the chain rule. …