Question

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

• A. \( 9e^{\pi/6} \)
• B. \( 3e^{\pi/6} \)
• C. \( 3e^{\pi/2} \)
• D. \( 9e^{\pi/2} \)

Answer 1

The Correct Option is D

We are given the equation \( \log_e y = 3 \sin^{-1} x \). We need to find \( (1 - x^2) y'' - xy' \) at \( x = \frac{1}{2} \).

Step 1: Differentiate \( y \)

Start by differentiating \( y = e^{3 \sin^{-1} x} \):

\[ \frac{1}{y} \cdot y' = 3 \cdot \frac{1}{\sqrt{1 - x^2}} \implies y' = \frac{3y}{\sqrt{1 - x^2}} \]

At \( x = \frac{1}{2} \), we get:

\[ y' = \frac{3e^{\frac{\pi}{6}}}{\sqrt{3}} = \frac{2\sqrt{3}e^{\frac{\pi}{6}}}{3} \]

Step 2: Differentiate \( y' \) to find \( y'' \)

Now differentiate \( y' \) to get \( y'' \):

\[ y'' = \frac{d}{dx} \left( \frac{3y}{\sqrt{1 - x^2}} \right) \]

Using the quotient rule, we get:

\[ y'' = 3 \left( \frac{\sqrt{1 - x^2} \cdot y' - y \cdot \left( -\frac{x}{\sqrt{1 - x^2}} \right)}{(1 - x^2)} \right) \]

Substitute the expression for \( y' \):

Step 3: Substitute \( x = \frac{1}{2} \)

At \( x = \frac{1}{2} \), we substitute values for \( y \) and \( y' \):

\[ (1 - x^2) y'' = 3 \left( 3e^{\frac{\pi}{6}} + \frac{e^{\frac{\pi}{6}}}{\sqrt{3}} \right) = 3e^{\frac{\pi}{6}} \left( 3 + \frac{1}{\sqrt{3}} \right) \]

Now, calculate \( (1 - x^2) y'' - xy' \):

\[ (1 - x^2) y'' - xy' = 3e^{\frac{\pi}{6}} \left( 3 + \frac{1}{\sqrt{3}} \right) - \frac{1}{2} \times \frac{2\sqrt{3}e^{\frac{\pi}{6}}}{3} \]

After simplifying:

\[ (1 - x^2) y'' - xy' = 9e^{\frac{\pi}{2}} \]

Thus, the correct answer is:

\( \boxed{9e^{\frac{\pi}{2}}} \)