If \(x^2y^3 = 32\), then the value of \(\displaystyle\frac{dy}{dx}\) at the point (2, 2) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle -\frac{2}{3}\)
The point lies on the curve, since \(2^2 \times 2^3 = 4 \times 8 = 32\). By implicit differentiation, using the product rule on the left:
\( \displaystyle 2x\,y^3 + x^2 \times 3y^2\frac{dy}{dx} = 0 \)
\( \displaystyle \frac{dy}{dx} = -\frac{2xy^3}{3x^2y^2} = -\frac{2y}{3x} \)
At (2, 2):
\( \displaystyle \frac{dy}{dx} = -\frac{2 \times 2}{3 \times 2} = -\frac{2}{3} \)
Why the other options are incorrect
Option (2) \(\displaystyle -\frac{3}{2}\) — Incorrect. The powers 2 and 3 have been interchanged: this is \(-\frac{3y}{2x}\).
Option (3) \(\displaystyle \frac{2}{3}\) — Incorrect. The minus sign is lost; the product \(x^2y^3\) is fixed, so y must fall as x rises.
Option (4) \(-1\) — Incorrect. This is the slope of the curve xy = 4 at (2, 2), where \(\frac{dy}{dx} = -\frac{y}{x}\); the powers 2 and 3 change the slope.