Question:hard

If $y = e^{a \sin^{-1} x}$, then $(1 - x^2) y_2 - x y_1 =$

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For functions of the form $y = e^{a f(x)}$, squaring and cross-multiplying the first derivative helps to eliminate the radical term before doing the second derivative.
Updated On: Jun 3, 2026
  • $a^2 y$
  • $-a^2 y$
  • $a y$
  • $-a y$
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The Correct Option is A

Solution and Explanation

Step 1: Look at the goal.
We start with $y = e^{a\sin^{-1}x}$ and must find the value of $(1 - x^2)y_2 - x y_1$, where $y_1$ is the first derivative and $y_2$ the second. We will differentiate step by step.

Step 2: Differentiate once.
Using the chain rule, the derivative of $e^{a\sin^{-1}x}$ brings out $a$ times the derivative of $\sin^{-1}x$, which is $\frac{1}{\sqrt{1-x^2}}$.
\[ y_1 = e^{a\sin^{-1}x}\cdot\frac{a}{\sqrt{1-x^2}} = \frac{a\,y}{\sqrt{1-x^2}} \]

Step 3: Remove the square root.
Multiply both sides by $\sqrt{1-x^2}$ so the root is gone.
\[ y_1\sqrt{1-x^2} = a\,y \]

Step 4: Square both sides.
Squaring clears the root completely and makes differentiating easier.
\[ y_1^2(1 - x^2) = a^2 y^2 \]

Step 5: Differentiate again.
Differentiate every term with respect to $x$, using the product rule on the left.
\[ 2y_1 y_2(1 - x^2) + y_1^2(-2x) = a^2(2y\,y_1) \]

Step 6: Cancel the common factor.
Every term has a factor $2y_1$. Divide it out (it is not zero).
\[ (1 - x^2)y_2 - x y_1 = a^2 y \] So the value asked for is $a^2 y$.
\[ \boxed{a^2 y} \]
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