Question

if (2xy^{2}-3x^{2}y = 6x), then (\frac{dy}{dx}=)
a) (\frac{6}{1y - 6x})
b) (\frac{2y^{2}-2xy - 6}{3x^{2}})
c) (\frac{-2y^{2}+6xy - 6}{4xy - 3x^{2}})
d) (\frac{-2y^{2}+6xy + 3x^{2}+6}{4xy})

Explanation:

Step1: Differentiate both sides with respect to \(x\)

Differentiate \(2xy^{2}-3x^{2}y = 6x\) using product - rule \((uv)^\prime=u^\prime v + uv^\prime\).
For the left - hand side:
The derivative of \(2xy^{2}\) is \(2y^{2}+4xy\frac{dy}{dx}\) (using product rule where \(u = 2x\), \(v = y^{2}\), \(u^\prime=2\), \(v^\prime = 2y\frac{dy}{dx}\)).
The derivative of \(-3x^{2}y\) is \(-6xy-3x^{2}\frac{dy}{dx}\) (using product rule where \(u=-3x^{2}\), \(v = y\), \(u^\prime=-6x\), \(v^\prime=\frac{dy}{dx}\)).
The derivative of the right - hand side \(6x\) is \(6\).
So, \(2y^{2}+4xy\frac{dy}{dx}-6xy - 3x^{2}\frac{dy}{dx}=6\).

Step2: Isolate \(\frac{dy}{dx}\)

Group the terms with \(\frac{dy}{dx}\) on one side:
\(4xy\frac{dy}{dx}-3x^{2}\frac{dy}{dx}=6 + 6xy-2y^{2}\).
Factor out \(\frac{dy}{dx}\):
\(\frac{dy}{dx}(4xy - 3x^{2})=6 + 6xy-2y^{2}\).
Then \(\frac{dy}{dx}=\frac{-2y^{2}+6xy + 6}{4xy-3x^{2}}\).

Answer:

D. \(\frac{-2y^{2}+6xy + 6}{4xy-3x^{2}}\)