Question

If \(x^3+y^3 = 6\) and \(\frac{d^2y}{dx^2}\cdot \frac{d^2x}{dy^2} = \frac{m}{(xy)^n}\), where \(m,n\in R\) then \(\frac{m}{n} =\)

• A. \(36\)
• B. \(9\)
• C. \(6\)
• D. \(4\)

Hint:

Find y double prime and x double prime separately, then multiply.

Answer 1

The Correct Option is A

Step 1: Use implicit differentiation twice using a shortcut:
For $x^3+y^3=6$, $\dfrac{dy}{dx}=-\dfrac{x^2}{y^2}$ and $\dfrac{dx}{dy}=-\dfrac{y^2}{x^2}$.

Step 2: Differentiate dx/dy w.r.t. y:
$\dfrac{d^2x}{dy^2}=-\dfrac{2yx^2-y^2\cdot2x\,x'}{x^4}$ with $x'=-y^2/x^2$, which simplifies to $-\dfrac{2y(x^3+y^3)}{x^5}=-\dfrac{12y}{x^5}$.

Step 3: Combine with d2y/dx2 = -12x/y^5:
The product is $\dfrac{144}{x^4y^4}$, so $m=144$, $n=4$ and $m/n=36$, option A.

Final Answer:
The product equals 144/(xy)^4, so m/n = 144/4 = 36. \[ \boxed{\text{(A) }36} \]