Differentiating exponential functions

[calculusthis]

k guys.. i'd like to try to figure out how to get to the derrivatives on my own so really just looking for a right or wrong.
1. f(x) = e^(3x^2+x) find f'(2)

f'(x) = e^(3x^2+x) * (6x+1)

f'(2)= e^(3(2)^2+2) * [6(2)+1]
f'(2)=13e^14 ... correct?2) find slope of the tangent to the function f(x) = 2^(x^2+3x) when x=3

f'(x) = (2x+3) 2^(x^2+3x) ln2

f'(3)= [2(3)+3] 2^[3^2+(3)(3)] ln 2
f'(3) = 1 635 339.37 ... correct?3) y = xe^(2x)
y' = 2x^2e^(2x) + e^(2x) ... correct?4) f(x) = 3^x + x^3
f'(x= 3^x ln 3 + 3x^2 ... correct?5) y= e^-(2x+5)
y' = e^(-2x-5) * (-2) ... correct?6) y = e^(3x^2-5x+7)
y' = e^(3x^2-5x+7) * (6x-5) ... correct?If anyone can quickly tell me if these are differentiated correctly it would be very helpful. I'm new to doing this type of work and find myself second guessing most my results.
thanx for any help.

 

[SammyS]

They look fine.

For #3, you could factor out e^2x.

Added in Edit:

Well ... Dick is right, there is a problem with #3.

 

[Dick]

I would check (3) again.

 

[calculusthis]

thanx guys... did you mean there's a problem with 3 other than i didnt factor out the e^2x?

 

[calculusthis]

I've been trying to work 3 differently but keep coming up with the same.. could someone show me where I'm going wrong please?

 

[Dick]

I've been trying to work 3 differently but keep coming up with the same.. could someone show me where I'm going wrong please?

Your first term in 3 should be x time the derivative of e^(2x). What's the derivative of e^(2x)?

 

[calculusthis]

the derivative of e^(2x) is... e^(2x)*(2x)..correct?

 

[calculusthis]

ohh wait i see.. just 2 not 2x lol

 

[calculusthis]

so the final answer should be y'=e^(2x)*(2x+1) ...?

 

[Dick]

ohh wait i see.. just 2 not 2x lol

You got it.

 

[calculusthis]

thanx a lot. ... I have another question about graphing.. If I'm given a graph of f'(x).. and it asks me to find a possible graph of f(x) and f''(x)... should I know the shape of f''(x) based off of f'(x) or am I just finding a possible graph for that as well?

 

[Dick]

To make a sketch of f''(x) just think about what the derivative of f'(x) should look like. f''(x) is just the slope of the curve f'(x). Right? To sketch a graph of f(x) think about what the integral of f'(x) would look like.

 

[calculusthis]

oh ok. So if I'm just given a graph and representing the function and not the equation of the function I'm just showing what the graphs should look like. Not what they actually look like.
thanx dick... you know, its not true what they say about you... you really don't like up to the name lol

 

[Dick]

oh ok. So if I'm just given a graph and representing the function and not the equation of the function I'm just showing what the graphs should look like. Not what they actually look like.
thanx dick... you know, its not true what they say about you... you really don't like up to the name lol

Well, your sketch of the graph of f''(x) should look pretty close to the graph you would get if you were given an equation. Where f'(x) is increasing, f''(x) better be positive, etc. The sketch of f(x) has a little more variable. Since integration give you a +C arbitrary constant you can shift the graph up or down. You are very welcome. And thanks for the compliment :).

 

[calculusthis]

ok... I'm going to add some pics.. could you let me know if there's anything i should change?

 

[calculusthis]

wait the last one is the wrong pic...

 

[calculusthis]

this is h''(x)

 

[Dick]

wait the last one is the wrong pic...

If you were going to multiply the last one by (-1), then everything looks ok. And you can shift the first one by any constant up or down, and it would still be ok.

 

[Dick]

this is h''(x)

Yes, ok!

 

[calculusthis]

well the first one is f(x) and it is given. the other two are the ones its asking for...

thanx again dick