- #1

- 1,291

- 0

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:

[tex]y=e^{-x^2}[/tex]

When I assign [tex]u=e^{-x}[/tex] and [tex]y=u^2[/tex] 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.

[tex]y=e^{-x^2}[/tex]

[tex]u=e^{-x} -----> \frac{du}{dx}=-e^{-x}[/tex]

[tex]y=u^2 ---------> \frac{dy}{du}=2u[/tex]

This shows out that the chain is consistent.

[tex]\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}

[tex]=2u(-e^{-x})[/tex]

[tex]=2(e^{-x})(-e^{-x})[/tex]

[tex]=-2e^{-2x}[/tex]

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

[tex]=-2x^{-x^2}[/tex]

[tex]y=e^{-x^2}[/tex]

When I assign [tex]u=e^{-x}[/tex] and [tex]y=u^2[/tex] 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.

[tex]y=e^{-x^2}[/tex]

[tex]u=e^{-x} -----> \frac{du}{dx}=-e^{-x}[/tex]

[tex]y=u^2 ---------> \frac{dy}{du}=2u[/tex]

This shows out that the chain is consistent.

[tex]\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}

[tex]=2u(-e^{-x})[/tex]

[tex]=2(e^{-x})(-e^{-x})[/tex]

[tex]=-2e^{-2x}[/tex]

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

[tex]=-2x^{-x^2}[/tex]

Last edited: