the tangent to the curve x^3+x^2y+4y=1 at the point (3,-2) has slope
a)-3 b)-23/9 c)-27/13 d)-11/9 e)-15/13

Question

the tangent to the curve x^3+x^2y+4y=1 at the point (3,-2) has slope
a)-3 b)-23/9 c)-27/13 d)-11/9 e)-15/13

loser66 · Answer

take derivative and plug the point in.

anonymous · Answer

The derivative can be computed in a few ways. I'll suggest two.

One method is via implicit differentiation, which yields
$$3x^2+2xy+x^2\frac{dy}{dx}+4\frac{dy}{dx}=0$$
From here you can solve for $\dfrac{dy}{dx}$ and plug in $x=3$ and $y=-2$.

A slightly different way: Separate your variables to solve for $y$, then take the derivative right away.
$$\begin{align*}
x^3+x^2y+4y&=1\\
(x^2+4)y&=1-x^3\\
y&=\frac{1-x^2}{x^2+4}\\
\frac{dy}{dx}&=\cdots
\end{align*}$$