Question

1. the slope of the tangent to the curve $y^{3}x + y^{2}x^{2} = 6$ at the point $(2,1)$ is

a) $-\frac{3}{2}$
b) $-1$
c) $-\frac{5}{14}$
d) $-\frac{3}{14}$
e) $0$

Explanation:

Step1: Differentiate the equation implicitly

Differentiate \(y^{3}x + y^{2}x^{2}=6\) with respect to \(x\) using the product rule \((uv)^\prime = u^\prime v+uv^\prime\).
For \(y^{3}x\), its derivative is \(3y^{2}\frac{dy}{dx}x + y^{3}\).
For \(y^{2}x^{2}\), its derivative is \(2y\frac{dy}{dx}x^{2}+2xy^{2}\).
So, \(3xy^{2}\frac{dy}{dx}+y^{3}+2x^{2}y\frac{dy}{dx}+2xy^{2}=0\).

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

Group the terms with \(\frac{dy}{dx}\): \(\frac{dy}{dx}(3xy^{2}+2x^{2}y)=-(y^{3} + 2xy^{2})\).
Then \(\frac{dy}{dx}=\frac{-(y^{3}+2xy^{2})}{3xy^{2}+2x^{2}y}\).

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

Substitute \(x = 2\) and \(y = 1\) into \(\frac{dy}{dx}\):

$$
LATEXBLOCK0
$$

Answer:

C. \(-\frac{5}{14}\)