# Homework Help: Taking the derivative of e^3x

1. Jan 11, 2009

### sonofjohn

If f(x) is the function given by f(x) = e3x + 1, at what value of x is the slope of the tangent line to f(x) equal to 2?

I thought the derivative of e3x would be 3e3x because of the chain rule, but it doesn't appear to be correct. I know that ex is just ex, so is e3x just e3x?

2. Jan 11, 2009

### Dick

You were right the first time. You need the chain rule. (e^(3x))'=3*e^(3x).

3. Jan 11, 2009

### tiny-tim

Hi sonofjohn!
It is correct … who says it isn't?

4. Jan 11, 2009

### sonofjohn

Ahh thank you very much!

5. Jan 11, 2009

### sonofjohn

The region bounded by y = e^x, y = 1, and x = 2 s rotated about the x-axis. The volume of the solid generated is given by the integral_____________.

So this is volume problem and a disc/ washer method formula should work best. I was going to set the problem as:

pi(antider)from 0-2 of (e^x-1)^2(1-1)^2 but I don't believe I will come out with 1-1 and 0 as the volume.

6. Jan 11, 2009

### Dick

You are rotating around the x-axis. At a given value of x what is the outer and inner radius of the washer?

7. Jan 11, 2009

### sonofjohn

0 and 2, so they would be my bounds so would I rather set the problem up as, pi*antiderv*(e^x-1)^2

8. Jan 11, 2009

### tiny-tim

Hi sonofjohn!
(have a pi: π )

I don't actually understand all of your (e^x-1)^2(1-1)^2 …

but you have a π(ex - 1)2, which is not the area of anything, is it?

9. Jan 11, 2009

### Dick

No, at say x=1 what is the inner and outer radius of the washer? What about at a general value of x? 0 and 2 are fine for the limits on the x integration. Now you just want to find the area of the washer. Correctly, this time. It's not pi*(e^x-1)^2.

10. Jan 11, 2009

### sonofjohn

Ok so I should find the area of the washer. The area of the washer should be defined as e^x - 1 Also since I am using the dish washer formula, I should square both parts of the integration thus yielding:

pi(anitderivative)(e^2x -1)

11. Jan 11, 2009

### Dick

12. Jan 11, 2009

### sonofjohn

Thanks for the clarification!

13. Jan 11, 2009

### sonofjohn

The 3rd problem on this page is difficult for me to understand. I don't understand what the x stands for when they are talking about the bounds in terms of integration. Do they mean the x-axis on the graph or possible x = 2 where at the x intercept?

14. Jan 11, 2009

### Dick

x is a point between 0 and 4. Any one. What you might notice is that G(x) is the area under the curve from 0 to x. H(x) is NEGATIVE of the area under the curve from x to 2 (because the integral is from 2 to x instead of from x to 2). Might this tell you something about G(x)-H(x)?

Last edited: Jan 11, 2009
15. Jan 11, 2009

### sonofjohn

So then the answer must be g(x) = h(x) - 2, because h(x) is positive and g(x) is negative. g(x) is also always going to be two less, because it goes from any point to 0, and h(x) only goes to 2.

16. Jan 11, 2009

### Dick

I can't really agree with you there, sonofjohn. Try this. What are G(1) and H(1)? Work them out from the picture.

17. Jan 11, 2009

### sonofjohn

I see now that (d) cannot work. Subtracting 2 everytime from h will not yield an equal integral. Now I would like to say that g(x) = h(x+2) would work, but it doesn't seem plausible past h(1). Could G'(x) = H'(x+2) work? I don't even understand what it means.

18. Jan 11, 2009

### Dick

You didn't answer my last question. What are G(1) and H(1)? That should let you eliminate some possibilities.

19. Jan 11, 2009

### sonofjohn

g(1) is -2 and h(1) is 1 so a,d,e cannot be the right answer .

Last edited: Jan 11, 2009
20. Jan 11, 2009

### Dick

How did you get that? I get G(1)=2 and H(1)=(-1).