#1

Hey Pit, I was just wondering who could answer a quick AP Calculus problem.

It has to do with Derivatives of Exponential functions; I know that the derivative of (e^x)=(e^x) , but I don't know what to do with the coefficients and such, when dealing with more complex problems.

Basically, I'd like to know how to go about doing a problem like:

Find an algebraic expression for f'(x) given f(x)=(e^5)-4e^x

Any help is appreciated.

(by the way, it's hard to see the f prime symbol with this font, but the problem is finding the derivative of the function).

Sorry, mistyped the first time.

It has to do with Derivatives of Exponential functions; I know that the derivative of (e^x)=(e^x) , but I don't know what to do with the coefficients and such, when dealing with more complex problems.

Basically, I'd like to know how to go about doing a problem like:

Find an algebraic expression for f'(x) given f(x)=(e^5)-4e^x

Any help is appreciated.

(by the way, it's hard to see the f prime symbol with this font, but the problem is finding the derivative of the function).

Sorry, mistyped the first time.

*Last edited by manmanman133 at Nov 9, 2008,*

#2

bump?

#3

There's no x. It's a constant. Should be 0.

#4

get your hands on a TI-89, and all your problems will go away.....

#5

there is no x in that expression so the derrivative is zero

#6

there is no x in that expression so the derrivative is zero

Sorry, a typo, it's corrected now.

#7

math thread

#8

The new one's derivative is -4e^x.

#9

derivitave will just be -4e^x , it doesnt change

#10

e^5 is a constant, so it turns into 0-int(4e^x)

4 is constant, so you get -4int(e^x)

which is -4e^x

4 is constant, so you get -4int(e^x)

which is -4e^x

#11

Thanks all, I've got it now.

#12

He's not integrating, but it's right nonetheless.

#13

-4x(e^[x-1])

#14

No no no no and no to the user above me...

This is how you do it...

f(x)=e^u

f '(x)=u'e^u

So for your problem...

f(x)=(e^5)-4e^x

(e^5) -> u=5 and u' =0

So 0(e^5)=0

-4e^x -> u=x and u' =1

So 1(-4e^x)=-4(e^x)

Therefore the answer is 0-4(e^x)

This is how you do it...

f(x)=e^u

f '(x)=u'e^u

So for your problem...

f(x)=(e^5)-4e^x

(e^5) -> u=5 and u' =0

So 0(e^5)=0

-4e^x -> u=x and u' =1

So 1(-4e^x)=-4(e^x)

Therefore the answer is 0-4(e^x)

*Last edited by MetropolisPt3 at Nov 9, 2008,*

#15

I used to think that tabs was the closest thing to math the Pit could do

lol

lol