If $x^2 +xy =10$ then when $x=2$ what is $\frac{\mathrm dy}{\mathrm dx}$?
If $x^2 +xy =10$ then when $x=2$ what is $\frac{\mathrm dy}{\mathrm dx}$?
I solved for $y=3$ before I did the product rule and i'm not sure if that was the correct way to approach it.
$\endgroup$ 24 Answers
$\begingroup$If $x^2+xy=10$, then $y=\frac{10}{x}-x$ (for $x\neq 0$). So $\frac{dy}{dx}=-\frac{10}{x^2}-1$.
This evaluated at $x=2$ is $\boxed{-\dfrac{7}{2}}$.
$\endgroup$ $\begingroup$Use implicit differentiation
$$ x^2 + xy = 10 \iff 2x + y + xy' = 0 \iff y' = \frac{-y-2x}{x}$$
When $x = 2$, $y = 3$. hence
$$y'(2,3) = \frac{-3-2(2)}{2} = \frac{-7}{2}$$
$\endgroup$ $\begingroup$$x^2+xy=10$ then when $x=2$ what is $dy/dx$?
$$x^2+xy=10\\ 4+2y=10\\ y=3$$
$$2x+x dy/dx + y=0\\ 4+2 dy/dx+y=0\\ 4+2 dy/dx+3=0$$
$$2 dy/dx=-7\\ dy/dx=-7/2$$
$\endgroup$ $\begingroup$You should use implicit differentiation here, since first solving for $y$ is meaningless (remember that the derivative is a property of functions, not of single numbers).
Thus, taking the derivative from both sides and using the product rule you get: $$2x + y + x\frac{dy}{dx}=0$$ So that: $$\frac{dy}{dx}=-\frac{2x+y}{x}$$
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