If log y = 3 sin⁻¹x, Then (1 − x²)y″ − xy′ at x = 1/2 Is Equal To

JEE Main 2024 — 9 April, Shift 2. Previous Year Question.

Problem

If $\log_e y = 3\sin^{-1}x$, then $(1-x^2)y”-xy’$ at $x=\dfrac12$ is equal to:

Key insight. Differentiating an implicit relation like $\log y=3\sin^{-1}x$ twice, and substituting the first derivative back into the second wherever possible, causes the messy square-root terms to cancel completely — leaving the whole expression $(1-x^2)y”-xy’$ as a clean constant multiple of $y$ itself.

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Approach

Differentiate $\log y=3\sin^{-1}x$ implicitly to get $y’$ in terms of $y$ and $x$. Rearrange this into the form $\sqrt{1-x^2}\,y’=3y$, then differentiate that form again — this avoids having to differentiate an explicit fraction with a square root in the denominator. Substituting the original expression for $y’$ back in at the right moment causes the square roots to cancel, leaving $(1-x^2)y”-xy’$ as a pure multiple of $y$. Finally, evaluate $y$ at $x=\dfrac12$ and substitute.

Solution

Step 1 — Differentiate log y = 3 sin⁻¹x implicitly

$$\frac1y\cdot y’ = \frac3{\sqrt{1-x^2}} \implies y’ = \frac{3y}{\sqrt{1-x^2}}$$

Step 2 — Rewrite to avoid a square root in the denominator

$$\sqrt{1-x^2}\,y’ = 3y$$

Step 3 — Differentiate both sides again

Using the product rule on the left (with $\dfrac{d}{dx}\sqrt{1-x^2}=\dfrac{-x}{\sqrt{1-x^2}}$):

$$\frac{-x}{\sqrt{1-x^2}}\,y’ + \sqrt{1-x^2}\,y” = 3y’$$

Step 4 — Multiply through by √(1−x²)

$$-x\,y’ + (1-x^2)y” = 3y’\sqrt{1-x^2}$$

$$(1-x^2)y”-xy’ = 3y’\sqrt{1-x^2}$$

Step 5 — Substitute y’ back in using Step 1

Since $y’=\dfrac{3y}{\sqrt{1-x^2}}$:

$$3y’\sqrt{1-x^2} = 3\left(\frac{3y}{\sqrt{1-x^2}}\right)\sqrt{1-x^2} = 9y$$

So:

$$(1-x^2)y”-xy’ = 9y$$

Step 6 — Evaluate y at x = 1/2

$$\log_e y = 3\sin^{-1}\left(\frac12\right) = 3\times\frac{\pi}6 = \frac{\pi}2$$

$$y = e^{\pi/2}$$

Step 7 — Substitute into the result

$$(1-x^2)y”-xy’ \Big|_{x=1/2} = 9y = 9e^{\pi/2}$$

Answer

Common mistakes

• Differentiating the explicit form $y’=\dfrac{3y}{\sqrt{1-x^2}}$ directly using the quotient rule. This works but produces a much messier expression than rearranging first to $\sqrt{1-x^2}\,y’=3y$, which is specifically shaped to simplify cleanly once differentiated and recombined.
• Forgetting to substitute $y’$ back in during Step 5. Without this substitution, the expression $3y’\sqrt{1-x^2}$ still contains $y’$ and $\sqrt{1-x^2}$ separately — recognising that these combine (via Step 1’s relation) to eliminate the square root entirely is what turns the answer into a clean multiple of $y$.
• Evaluating $\sin^{-1}(1/2)$ incorrectly. $\sin^{-1}\left(\dfrac12\right)=\dfrac{\pi}6$ (30°), not $\dfrac{\pi}3$ (60°, which is $\cos^{-1}(1/2)$) — mixing these up changes the final exponent from $\pi/2$ to $\pi$.

Practise next

• If $\log_e y=2\sin^{-1}x$, find $(1-x^2)y”-xy’$ at $x=\dfrac12$.