Question:medium

If \( y = \cos^{-1} x \), find \( \frac{d^2y}{dx^2} \) in terms of \( y \).

Show Hint

For second derivatives in inverse trig functions, always convert to \(x = f(y)\) first. It makes chain rule application much cleaner and avoids messy algebra.
Updated On: May 29, 2026
  • \(-\csc^2 y \cot y\)
  • \(\csc^2 y \cot y\)
  • \(-\csc y \cot^2 y\)
  • \(\csc y \cot^2 y\)
Show Solution

The Correct Option is A

Solution and Explanation

Step 1 : Understanding the Question:
The objective of this problem is to calculate the second-order derivative of the inverse cosine function, $y = \cos^{-1} x$. However, the specific constraint is that the final expression must be represented purely as a function of $y$, without any occurrence of the variable $x$. This requires a deeper manipulation of trigonometric identities and a clever use of the chain rule to transition from $x$-based differentiation to $y$-based expressions. This is a common requirement in calculus to facilitate further analysis of the rate of change in terms of the output variable.
Step 2 : Key Formulas and approach:
We first transform the inverse trigonometric equation into its direct trigonometric counterpart: $x = \cos y$. This allows us to differentiate implicitly with respect to $x$. During this process, we utilize the chain rule, which states that $\frac{d}{dx}[f(y)] = f'(y) \frac{dy}{dx}$. For the second derivative, we will differentiate the first derivative again, remembering to multiply by $dy/dx$ whenever we differentiate a term containing $y$.
Key derivatives:
1. $\frac{d}{dy}(\cos y) = -\sin y$
2. $\frac{d}{dy}(\csc y) = -\csc y \cot y$
Step 3 : Detailed Explanation:

We rewrite the given function $y = \cos^{-1} x$ as $x = \cos y$.

Differentiating both sides with respect to $x$ gives: $1 = -\sin y \cdot \frac{dy}{dx}$.

Rearranging for the first derivative, we get $\frac{dy}{dx} = -\frac{1}{\sin y}$, which simplifies to $\frac{dy}{dx} = -\csc y$.

To find the second derivative, we differentiate $dy/dx$ with respect to $x$: $\frac{d^2y}{dx^2} = \frac{d}{dx}(-\csc y)$.

Applying the chain rule, we differentiate with respect to $y$ and then multiply by $dy/dx$: $\frac{d^2y}{dx^2} = \frac{d}{dy}(-\csc y) \cdot \frac{dy}{dx}$.

The derivative of $-\csc y$ with respect to $y$ is $-(-\csc y \cot y) = \csc y \cot y$.

Substituting the value of $dy/dx = -\csc y$ back into the equation, we get $\frac{d^2y}{dx^2} = (\csc y \cot y) \cdot (-\csc y)$.

Multiplying these terms results in the final expression: $\frac{d^2y}{dx^2} = -\csc^2 y \cot y$.

Step 4 : Final Answer:
The second-order derivative in terms of $y$ is $-\csc^2 y \cot y$.
Was this answer helpful?
0