Question

If \(y = sin^{-1}(\frac{25-x^2}{25+x^2})\), then \(y^'(1)\) is...

• A. \(-\frac{5}{13}\)
• B. \(\frac{5}{13}\)
• C. \(\frac{13}{5}\)
• D. \(\frac{2}{7}\)

Hint:

Substitute x = 5 tan(theta) to turn the expression into cos 2 theta.

Answer 1

The Correct Option is A

Step 1: Understand the concept
The expression \(\dfrac{25 - x^2}{25 + x^2}\) looks like \(\dfrac{1 - \tan^2\theta}{1 + \tan^2\theta} = \cos 2\theta\) with \(x = 5\tan\theta\).

Step 2: Simplify y
Then \(y = \sin^{-1}(\cos 2\theta) = \sin^{-1}\left[\sin\left(\frac{\pi}{2} - 2\theta\right)\right] = \frac{\pi}{2} - 2\theta\). This holds for \(0 \le \theta \le \frac{\pi}{2}\), which includes \(x = 1\). So
\[ y = \frac{\pi}{2} - 2\tan^{-1}\frac{x}{5} \]

Step 3: Differentiate
\[ y' = -2\cdot\frac{1}{1 + x^2/25}\cdot\frac{1}{5} = -\frac{10}{25 + x^2} \]

Step 4: Evaluate at x = 1
\(y'(1) = -\dfrac{10}{26} = -\dfrac{5}{13}\), option (A). Option (B) has the right size but the wrong sign.

Final Answer:
y' at 1 is -5/13. This is option (A). \[ \boxed{\text{(A) }-\frac{5}{13}} \]