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

6. the slope of the tangent to the curve $y^{3}x + y^{2}x^{2} = 6$ at the point $(2,1)$ is\na) $-\\frac{3}{2}$\nb) $-1$\nc) $-\\frac{5}{14}$\nd) $-\\frac{3}{14}$\ne) $0$

6. the slope of the tangent to the curve $y^{3}x + y^{2}x^{2} = 6$ at the point $(2,1)$ is\na) $-\\frac{3}{2}$\nb) $-1$\nc) $-\\frac{5}{14}$\nd) $-\\frac{3}{14}$\ne) $0$

Answer

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}): [ \begin{align*} \frac{dy}{dx}&=\frac{-(1^{3}+2\times2\times1^{2})}{3\times2\times1^{2}+2\times2^{2}\times1}\ &=\frac{-(1 + 4)}{6 + 8}\ &=\frac{-5}{14} \end{align*} ]

Answer:

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