Question

4x - x²y + y³ = 10
find the value of \frac{dy}{dx} at the point (1, 2).
choose 1 answer:
(a) \frac{2}{3}
(b) -4
(c) -1
(d) 0

Explanation:

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

Differentiate \(4x - x^{2}y + y^{3}\) term - by - term.
Using the sum/difference rule \((u\pm v\pm w)'=u'\pm v'\pm w'\), we have \(\frac{d}{dx}(4x)-\frac{d}{dx}(x^{2}y)+\frac{d}{dx}(y^{3})=\frac{d}{dx}(10)\).
The derivative of \(4x\) with respect to \(x\) is \(4\) (since \(\frac{d}{dx}(ax)=a\) for \(a = 4\)).
For \(\frac{d}{dx}(x^{2}y)\), use the product rule \((uv)' = u'v+uv'\), where \(u = x^{2}\), \(u'=2x\) and \(v = y\), \(v'=\frac{dy}{dx}\). So \(\frac{d}{dx}(x^{2}y)=2xy+x^{2}\frac{dy}{dx}\).
For \(\frac{d}{dx}(y^{3})\), use the chain rule. Let \(u = y\), then \(\frac{d}{dx}(y^{3})=3y^{2}\frac{dy}{dx}\).
The derivative of a constant \(10\) with respect to \(x\) is \(0\).
So, \(4-(2xy + x^{2}\frac{dy}{dx})+3y^{2}\frac{dy}{dx}=0\).

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

Expand the left - hand side: \(4-2xy - x^{2}\frac{dy}{dx}+3y^{2}\frac{dy}{dx}=0\).
Group the terms with \(\frac{dy}{dx}\) together: \((3y^{2}-x^{2})\frac{dy}{dx}=2xy - 4\).
Then \(\frac{dy}{dx}=\frac{2xy - 4}{3y^{2}-x^{2}}\).

Step3: Substitute \(x = 1\) and \(y = 2\)

Substitute \(x = 1\) and \(y = 2\) into \(\frac{dy}{dx}=\frac{2xy - 4}{3y^{2}-x^{2}}\).
\(\frac{dy}{dx}=\frac{2\times1\times2-4}{3\times2^{2}-1^{2}}=\frac{4 - 4}{12 - 1}=0\).

Answer:

D. \(0\)