Register to reply 
Taking the derivative of e^3xby sonofjohn
Tags: derivative 
Share this thread: 
#1
Jan1109, 04:45 PM

P: 76

If f(x) is the function given by f(x) = e^{3x} + 1, at what value of x is the slope of the tangent line to f(x) equal to 2?
I thought the derivative of e^{3x} would be 3e^{3x} because of the chain rule, but it doesn't appear to be correct. I know that e^{x} is just e^{x}, so is e^{3x} just e^{3x}? 


#2
Jan1109, 04:56 PM

Sci Advisor
HW Helper
Thanks
P: 25,251

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



#3
Jan1109, 04:56 PM

Sci Advisor
HW Helper
Thanks
P: 26,160

Hi sonofjohn!



#4
Jan1109, 04:58 PM

P: 76

Taking the derivative of e^3x
Ahh thank you very much!



#5
Jan1109, 05:05 PM

P: 76

The region bounded by y = e^x, y = 1, and x = 2 s rotated about the xaxis. 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 02 of (e^x1)^2(11)^2 but I don't believe I will come out with 11 and 0 as the volume. 


#6
Jan1109, 05:11 PM

Sci Advisor
HW Helper
Thanks
P: 25,251

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



#7
Jan1109, 05:16 PM

P: 76

0 and 2, so they would be my bounds so would I rather set the problem up as, pi*antiderv*(e^x1)^2



#8
Jan1109, 05:18 PM

Sci Advisor
HW Helper
Thanks
P: 26,160

Hi sonofjohn!
I don't actually understand all of your (e^x1)^2(11)^2 … but you have a π(e^{x}  1)^{2}, which is not the area of anything, is it? 


#9
Jan1109, 05:19 PM

Sci Advisor
HW Helper
Thanks
P: 25,251

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^x1)^2.



#10
Jan1109, 05:25 PM

P: 76

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
Jan1109, 05:29 PM

Sci Advisor
HW Helper
Thanks
P: 25,251

Ok. Yes. The area is pi*(outer radius^2inner radius^2).



#12
Jan1109, 05:32 PM

P: 76




#13
Jan1109, 06:08 PM

P: 76

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 xaxis on the graph or possible x = 2 where at the x intercept? 


#14
Jan1109, 06:29 PM

Sci Advisor
HW Helper
Thanks
P: 25,251

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)?



#15
Jan1109, 06:44 PM

P: 76

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
Jan1109, 06:50 PM

Sci Advisor
HW Helper
Thanks
P: 25,251




#17
Jan1109, 07:01 PM

P: 76

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
Jan1109, 07:07 PM

Sci Advisor
HW Helper
Thanks
P: 25,251

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



Register to reply 
Related Discussions  
Taking the derivative of a definite integral  Calculus  11  
Taking Physical Chemistry before formally taking Quantum Mechanics  Academic Guidance  17  
Replacing total derivative with partial derivative in Griffiths' book  Advanced Physics Homework  3  
Taking derivative  Calculus & Beyond Homework  2  
Taking the derivative  Calculus  11 