Question

2x^{3}-5xy - y^{2}=3
find \frac{dy}{dx}.
choose 1 answer:
a \frac{6x^{2}-5y}{5x + 2y}
b \frac{6x-2y}{5}
c \frac{6x^{2}-5y}{2y - 5x}

Explanation:

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

Differentiate \(2x^{3}-5xy - y^{2}=3\) term - by - term.
Using the power rule \(\frac{d}{dx}(x^{n})=nx^{n - 1}\), \(\frac{d}{dx}(2x^{3})=6x^{2}\).
For the term \(-5xy\), use the product rule \(\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}\), where \(u=-5x\) and \(v = y\). So \(\frac{d}{dx}(-5xy)=-5y-5x\frac{dy}{dx}\).
For the term \(-y^{2}\), use the chain rule \(\frac{d}{dx}(u^{n})=nu^{n - 1}\frac{du}{dx}\), where \(u = y\) and \(n = 2\). So \(\frac{d}{dx}(-y^{2})=-2y\frac{dy}{dx}\).
Differentiating the right - hand side \(\frac{d}{dx}(3)=0\).
The derivative of the left - hand side is \(6x^{2}-5y-5x\frac{dy}{dx}-2y\frac{dy}{dx}\), and the derivative of the right - hand side is \(0\). So \(6x^{2}-5y-5x\frac{dy}{dx}-2y\frac{dy}{dx}=0\).

Step2: Solve for \(\frac{dy}{dx}\)

Group the terms with \(\frac{dy}{dx}\) together:
\(-5x\frac{dy}{dx}-2y\frac{dy}{dx}=5y - 6x^{2}\).
Factor out \(\frac{dy}{dx}\): \(\frac{dy}{dx}(-5x - 2y)=5y - 6x^{2}\).
Then \(\frac{dy}{dx}=\frac{6x^{2}-5y}{5x + 2y}\).

Answer:

A. \(\frac{6x^{2}-5y}{5x + 2y}\)