Question

If \(y^{\frac{1}{m}}+y^{\frac{-1}{m}} = 2x\) , then \((x^2-1)y_1^2 =\)

• A. \(my^2\)
• B. \(m^2y^2\)
• C. \(m^2y\)
• D. \(my\)

Hint:

Solve for y^(1/m) and write y as a power of x + sqrt(x^2 - 1).

Answer 1

The Correct Option is B

Step 1: Implicit differentiation:
Differentiate $y^{1/m} + y^{-1/m} = 2x$: $\tfrac1m y^{1/m - 1}y_1 - \tfrac1m y^{-1/m - 1}y_1 = 2$.

Step 2: Factor:
Multiply the whole equation by $my$: $y_1\big(y^{1/m} - y^{-1/m}\big) = 2my$.

Step 3: Use squares:
$\big(y^{1/m} - y^{-1/m}\big)^2 = \big(y^{1/m}+y^{-1/m}\big)^2 - 4 = 4x^2 - 4$. Squaring: $y_1^2\cdot4(x^2-1) = 4m^2y^2$, so $(x^2-1)y_1^2 = m^2y^2$, option (B).

Final Answer:
The value is m squared y squared. \[ \boxed{\text{(B) }m^2y^2} \]