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: Direct differentiation
With $u = \dfrac{25 - x^2}{25 + x^2}$, we have $y' = \dfrac{u'}{\sqrt{1 - u^2}}$.

Step 2: Compute u'
$u' = \dfrac{-2x(25 + x^2) - 2x(25 - x^2)}{(25 + x^2)^2} = \dfrac{-100x}{(25+x^2)^2}$. And $1 - u^2 = \dfrac{(25+x^2)^2 - (25-x^2)^2}{(25+x^2)^2} = \dfrac{100x^2}{(25+x^2)^2}$.

Step 3: Combine
$y' = \dfrac{-100x}{(25+x^2)^2}\cdot\dfrac{25 + x^2}{10x} = -\dfrac{10}{25 + x^2}$ for $x > 0$.

Step 4: Evaluate
At $x = 1$, $y' = -\frac{10}{26} = -\frac{5}{13}$.

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