Find $\frac{dy}{dt}$ for the given x-values.

Question

A point moves along the curve of the given equation such that $\frac{dx}{dt}$ is 2 cm/s. Find $\frac{dy}{dt}$ for the given values of $x$.

$$y= \frac{1}{1+x^2};$$ $$x=-2, x=0, x=2$$

I've just started to learn about dealing with inverses in my calculus class and I have no idea what I'm doing. I don't even know how to start this problem and my homework is all word problems like this.

Answer 1

By the chain rule we have:

$$\frac{dy}{dt}=\frac{\partial y}{\partial x}\frac{dx}{dt}$$ So: $$\frac{dy}{dt}(x)=\frac{-2x}{(1+x^2)^2}*2=\frac{-4x}{(1+x^2)^2}$$ Filling in the $x$s we find: $$\frac{dy}{dt}(-2)=\frac{4}{25}\;\;\;\;\;\frac{dy}{dt}(0)=0\;\;\;\;\;\frac{dy}{dt}(2)=-\frac{4}{25}$$ ($\mathrm{cm}/\mathrm{s}$)

Answer 2

$$y(1+x^2)=1$$

$$y(2x \frac{\partial x}{\partial t})+\frac{\partial y}{\partial t}(1+x^2)=0$$

at each of the mentioned points, you can calculate $y$ and by a simple substitution of $x$ and $\frac{\partial x}{\partial t}=2 cm/s$ you can obtain $\frac{\partial y}{\partial t}$ easily.