1. ## Chain Rule clarification

I have to take the AP Calculus AB test tomorrow, so anyone who can clarify this for me quickly would be much, much appreciated.

I understand that when I perform the chain rule with a derivative, such as d/dx (2x+1)^3, I must take the derivative, yielding 3(2x+1)^2, and then I must multiply by the derivative of the inside, d/dx (2x+1) = 2, yielding 2*3(2x+1)^2 -> 6(2x+1)^2.

However, I am confused on this when I am using something other than a simple binomial. If I must evaluate d/dx (e^2x), would the derivative simply be (e^2x), or would I multiply that by the derivative of the exponent (d/dx 2x = 2), making it 2*(e^2x) -> (2e^2x)?

If the latter is true and I must multiply by the derivative of the exponent, then what do I do when I must take the antiderivative of it? For example, if I am given ∫(e^2x)dx, would the answer be (1/2)(e^2x)+C ?

Please correct anything I am mistaken on. Thank you.

2. Your reasoning is correct. Try the substitution $\displaystyle u=2x$ to make it even clearer.

$\displaystyle \frac{d}{dx}\left(e^u\right)=\frac{d}{du}(e^u) \cdot \frac{du}{dx}=e^u \cdot 2 = 2e^{2x}$

3. Right on dude. Looks like you got it to me. good luck on the exam.
$\displaystyle \frac{d}{dx}e^{2x}=2e^{2x}$

4. Thank you! =D

5. Originally Posted by Von Lauder
I have to take the AP Calculus AB test tomorrow, so anyone who can clarify this for me quickly would be much, much appreciated.

I understand that when I perform the chain rule with a derivative, such as d/dx (2x+1)^3, I must take the derivative, yielding 3(2x+1)^2, and then I must multiply by the derivative of the inside, d/dx (2x+1) = 2, yielding 2*3(2x+1)^2 -> 6(2x+1)^2.

However, I am confused on this when I am using something other than a simple binomial. If I must evaluate d/dx (e^2x), would the derivative simply be (e^2x), or would I multiply that by the derivative of the exponent (d/dx 2x = 2), making it 2*(e^2x) -> (2e^2x)?

correct

If the latter is true and I must multiply by the derivative of the exponent, then what do I do when I must take the antiderivative of it? For example, if I am given ∫(e^2x)dx, would the answer be (1/2)(e^2x)+C ?

correct

Please correct anything I am mistaken on. Thank you.
u a function of x ...

$\displaystyle \frac{d}{dx} [e^u] = e^u \cdot \frac{du}{dx}$

k a konstant ...

$\displaystyle \int e^{kx} \, dx = \frac{1}{k} e^{kx} + C$

6. Originally Posted by Von Lauder
I have to take the AP Calculus AB test tomorrow, so anyone who can clarify this for me quickly would be much, much appreciated.

I understand that when I perform the chain rule with a derivative, such as d/dx (2x+1)^3, I must take the derivative, yielding 3(2x+1)^2, and then I must multiply by the derivative of the inside, d/dx (2x+1) = 2, yielding 2*3(2x+1)^2 -> 6(2x+1)^2.

However, I am confused on this when I am using something other than a simple binomial. If I must evaluate d/dx (e^2x), would the derivative simply be (e^2x), or would I multiply that by the derivative of the exponent (d/dx 2x = 2), making it 2*(e^2x) -> (2e^2x)?
Yes, any time you have a "differentiation rule" for f(x) but have some function of x, say u(x), you must use the chain rule:
$\displaystyle \frac{df(u(x))}{dx}= \frac{df(u)}{du}\frac{du}{dx}$
In this particular example, u(x)= 2x so
$\displaystyle \frac{de^u}{dx}= \frac{de^u}{du}\frac{du}{dx}= \frac{e^u}{du}\frac{2x}{dx}= (e^u)(2)= 2e^{2x}$

A slightly more complicated example is:
$\displaystyle \frac{d e^{x^2}}{dx}$
Here $\displaystyle u(x)= x^2$ so
$\displaystyle \frac{d e^{x^2}}{dx}= \frac{de^u}{du}\frac{du}{dx}= \frac{de^u}{du}\frac{d(x^2)}{dx}= (e^u)(2x)= 2xe^{x^2}$

If the latter is true and I must multiply by the derivative of the exponent, then what do I do when I must take the antiderivative of it? For example, if I am given ∫(e^2x)dx, would the answer be (1/2)(e^2x)+C ?
That's exactly what "substitution" in integration is. To integrate \int e^{2x}dx[/tex], let u= 2x. then du= 2dx so (1/2)du= dx. Replacing 2x by u and dx by (1/2)du, the integral becomes
[tex]\int e^{2x}dx= \int e^u ((1/2)du)= (1/2)\int e^u du= (1/2)e^u+ C= (1/2)e^{2x}+ C[//math]
just as you say.

HOWEVER, integrating is harder than differentiating and this is one reason. If the integral were
$\displaystyle \int e^{x^2} dx$
we might try the substitution $\displaystyle u= x^2$. Then du= (1/2)x dx. If we do the same as before, 2du/x= dx and when we replace $\displaystyle x^2$ by u and dx by 2du/x we get
$\displaystyle \int 2 e^u/x du$
and still have x in the integral. We can't take it out of the integral as we did with the "2" in the first integral because it is a variable and not a constant. That is why
$\displaystyle \int e^{-x^2} dx$
important in the "Normal Probability Distribution", cannot be written in terms of elementary functions.

Notice that, even though it seems more difficult, the integral
$\displaystyle \int xe^{x^2}dx$
CAN be done easily. Using the substitution $\displaystyle u= x^2$ now, $\displaystyle du= 2x dx$, [tex] (1/2)du= x dx and the integral becomes
$\displaystyle (1/2)\int e^u du= (1/2)e^u+ C= (1/2)e^{x^2}+ C$

Please correct anything I am mistaken on. Thank you.

7. ## Possible Correction.

Originally Posted by HallsofIvy
HOWEVER, integrating is harder than differentiating and this is one reason. If the integral were
$\displaystyle \int e^{x^2} dx$
we might try the substitution $\displaystyle u= x^2$. Then du= (1/2)x dx. If we do the same as before, 2du/x= dx and when we replace $\displaystyle x^2$ by u and dx by 2du/x we get
$\displaystyle \int 2 e^u/x du$
and still have x in the integral. We can't take it out of the integral as we did with the "2" in the first integral because it is a variable and not a constant.
If $\displaystyle u=x^2$, then I think that $\displaystyle du = 2x dx$, not $\displaystyle du = \frac{1}{2}x dx$. I think what you meant was that $\displaystyle dx = \frac{1}{2x}du$.

Either way, you are stuck with two variables in the integral. I just thought I'd point that out. (And there is a good chance that you are right, and I am wrong by the way).