# Another chain rule question.

• Nov 7th 2010, 05:38 AM
Another chain rule question.
Can you also check this out for me;

$y=e^a$

where $a=-x^2/2$

1... $dy/dx = dy/du \cdot du/dx$

2...where $y = e ^-^u$ and $u= - x^2/2$

3.... $dy/du=e ^-^u$ and $du/dx= -x$ <<<< can you explain how step 2 for "u" this turns into -x

4.... therefore; $dy/du \cdot du/dx = -x e ^ - x^2/2$
• Nov 7th 2010, 05:53 AM
e^(i*pi)
Quote:

Can you also check this out for me;

$y=e^a$

where $a=-x^2/2$

1... $dy/dx = dy/du \cdot du/dx$

2...where $y = e ^-^u$ and $u= - x^2/2$

3.... $dy/du=e ^-^u$ and $du/dx= -x$ <<<< can you explain how step 2 for "u" this turns into -x

4.... therefore; $dy/du \cdot du/dx = -x e ^ - x^2/2$

You are using a instead of u. Minor difference but instead you need to find

1. $\dfrac{dy}{dx} = \dfrac{dy}{da} \cdot \dfrac{da}{dx}$

2. Aside from my above note about a then yes.

3. You need to use the chain rule. It's a tricky one to spot but if you imagine $e^{-a} = e^{-1 \cdot a}$ so the derivative is multiplied by -1. Thus $\dfrac{d}{da}e^{-a} = -e^{-a}$

For $\dfrac{da}{dx} = -x$ is true because you can write it as $-\dfrac{1}{2} \cdot x^2$. x^2 is simple enough but you must multiply by -1/2 which gives -x

4. Almost, because of the sign error in finding $\dfrac{da}{dx}$ your answer is a factor of -1 out.

$\dfrac{dy}{dx} = -x \cdot -e^{-x^2/2} = xe^{-x^2/2}$