Question about implicit differentiation, finding $\frac{dx}{dt}$

Suppose $xy=2$ and $\dfrac{dy}{dt}=2$. Find $\dfrac{dx}{dt}$ when $x=3$.

I don't know how to solve this. I know that if I'm differentiating with respect to time $t$, then the chain rule states that $\dfrac{d}{dt}(xy)=\dfrac{dy}{dt}x+\dfrac{dx}{dt}y$, but I don't know how to proceed from here...

Thank you.

• Replace y in terms of x. Also the derivative a constant is 0. Commented Oct 5, 2014 at 23:04

3 Answers

Since $xy=2$ you have that $$0=\frac{d{\bf{2}}}{dt}=x\frac{dy}{dt}+y\frac{dx}{dt}.$$ Now, you know $x=3$ (and so $y=2/3$) and $\frac{dy}{dt}=2.$ Thus, you can get $\frac{dx}{dt}.$

• OK, why isn't $\dfrac{d}{dt}(xy)=\dfrac{dy}{dt}x+\dfrac{dx}{dt}y$? why is it $\dfrac{d}{dt}(xy)=\dfrac{dy}{dt}x+\dfrac{dy}{dt}y$? thanks +1 Commented Oct 5, 2014 at 23:24
• Sorry. It was a typo. I have modified it. Thanks for your observation.
– mfl
Commented Oct 5, 2014 at 23:26

Hint: $$\frac{dx}{dt} = \frac{dx}{dy}\cdot \frac{dy}{dt}$$ You already have $\frac{dy}{dt}$ and can solve for $\frac{dx}{dy}$ by noting that $y = \frac{2}{x}$

$$xy=2$$

$$\implies \frac {d}{dx}(xy)=0$$

$$\implies y+x\frac {dy}{dx}=0$$

$$\implies \frac {dy}{dx}=-\frac {y}{x}$$

$$\implies \frac {dy}{dt} \cdot \frac {dt}{dx}=-\frac {2}{x^2}$$ {$$xy=2$$}

$$\implies \frac {dx}{dt}=\frac {-x^2}{2} \frac {dy}{dt}$$

$$\implies \frac {dx}{dt}=\frac {-3^2}{2} \cdot 2=-9$$