# When to assign a value to a multiple chain derivative?

I was doing my homework and I ran into a problem of a chain rule within a chain rule. When do I know what to assign a value? For example:

$$y=e^{-x^2}$$

When I assign $$u=e^{-x}$$ and $$y=u^2$$ I get a wrong value. According to cramster I was supposed to assign y = e^u and u=-x^2. But when am I supposed to know what value to assign? For clarification let me write out the steps my problem.

$$y=e^{-x^2}$$
$$u=e^{-x} -----> \frac{du}{dx}=-e^{-x}$$
$$y=u^2 ---------> \frac{dy}{du}=2u$$
This shows out that the chain is consistent.
$$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx} [tex]=2u(-e^{-x})$$
$$=2(e^{-x})(-e^{-x})$$
$$=-2e^{-2x}$$

But cramster has a different answer. Cramster assigned the values as y=e^u and u =-x^2 got an answer of:
$$=-2x^{-x^2}$$

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gb7nash
Homework Helper
When I assign $$u=e^{-x}$$ and $$y=u^2$$ I get a wrong value.

This is your problem. $u^{2}$ is not equal to $e^{-x^2}$. In fact, it's equal to $e^{-x}e^{-x} = e^{-2x}$

This is your problem. $u^{2}$ is not equal to $e^{-x^2}$. In fact, it's equal to $e^{-x}e^{-x} = e^{-2x}$

No I assign $$y=u^2$$. So then $$f'(y)=2u$$

Edit: I think there is a bit of a confusion. This is the answer I got-->$$e^{-x} = e^{-2x}$$. I think you confused it with my last line; the last line is the answer posted up on cramster. I edited my original post and clarified it.

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gb7nash
Homework Helper
No I assign $$y=u^2$$

Is this the same y as:

$$y=e^{-x^2}$$

?

Read my previous post I edited it, it might clarify things. Basically when there is a chain rule within a chain rule I assign a value. Where in this case it was:

$$y=u^2$$ and $$u=e^{-x}$$

Ohhhhh.. so in this case u^2 implies the function $$(e^{-x})^2$$ Where it is supposed to be only $$-x^2$$

That is what your saying right? Because I just noticed.

gb7nash
Homework Helper
Ohhhhh.. so in this case u^2 implies the function $$(e^{-x})^2$$ Where it is supposed to be only $$-x^2$$

That is what your saying right? Because I just noticed.

Correct