# Is d/dx e^-x = e^-x ?

1. Dec 9, 2008

### matttan

1st qs: I know that d/dx of e^x is e^x, but how bout negative power?(see example below)
Is d/dx e^-x = e^-x ?

and 2nd qs: If the power if not x but 3x + 4, y=e^(3x+4) then is d/dx e^(3x+4) = e^(3x+4) ?

2. Dec 9, 2008

### epenguin

You mean d(e^-x)/dx

Do you know what is df[g(x)]/dx in general?

3. Dec 9, 2008

### matttan

I will use other notation

If f(x)=e^-x then do f'(x)=e^-x ?

4. Dec 9, 2008

### matttan

I do not know, please enlight me.

5. Dec 9, 2008

### epenguin

In another notation do you know what [f(g(x))]' is ?

(And apart from calculation it is possible to visualise it.)

6. Dec 9, 2008

### matttan

Yup I think I know f(g(x)) but are u using this to explain d/dx e^-x = e^-x and before that, is d/dx e^-x = e^-x correct?

7. Dec 9, 2008

### aLiase

1. d(e^-x)/dx = -e^-x, use the chain rule to solve it.
2. Again, use the chain rule, take the derivative of e first then 3x+4

8. Dec 9, 2008

### epenguin

e-x is just a very simple example of f[g(x)] .

To be asking this question you must have done some calculus lessons before. It is a mistake to never go back to look at earlier lessons.

OK - what is f'(-x) in general??

9. Dec 9, 2008

### symbolipoint

For question #1, e-x=$$\frac{1}{e^x}$$
So you are looking for the derivative of a rational function.
EDIT: the formatting tool did not work. RightSide is supposed to be 1/(e^x)

Last edited by a moderator: Dec 12, 2008
10. Dec 9, 2008

### symbolipoint

Actually, I am not certain if my intended response in #9 is correct; I BELIEVE it is correct only because I could understand the meaning of negative exponent. (For me, been a long time since Calculus)

11. Dec 10, 2008

### epenguin

Sorry, I meant what is [f(-x)]' in general?

12. Dec 10, 2008

### matt_crouch

d/dx of e^-x = -e^-x

if all else fails or you want further explaination try using the chain rule
when e^-x the coefficiant of x is -1 right? so you time e by -1 (not very clear i know sorry)

13. Dec 10, 2008

### robert Ihnot

The way I like to look at that problem is: Let Y = e^(-x), then InY = -x. While we know that the derivative of -x = -1, it is also necessary to know the derivative of InY, which I leave to others.

(So that this way is actually a little advanced and is part of differential equations.)

14. Dec 10, 2008

### HallsofIvy

Staff Emeritus
What people have been trying to tell you is to use the chain rule.

The derivative of f(g)(x) is [f(g)(x)]'= f'(g)(x)g'(x) or df(g)(x)/dx= (d(f(g))/dg)(dg/dx). Now to differentiate e-x so that g(x)= -x. Then f(g)= eg and, apparently, you know that d eg/dg= eg. What is dg/dx= d(-x)/dx?
Multiply those together.

15. Dec 11, 2008

### epenguin

Yes, you need to go back to basics.

Plus get the habit of checking for what sounds right and reasonable, plausible. E.g. you probably know ex is always positive and always increases with x. I.e. d(ex)/dx is always positive.

So e-x which is 1/ex, must be positive and always decreasing with increasing x. i.e. d(e-x)/dx is always negative. One way to know your initial conjecture had to be wrong.

Last edited: Dec 11, 2008
16. Dec 12, 2008

### matt_crouch

How do you differentiate

e^3x (sinx+2cosx)

i understand that you use the product rule but its the very last bit of the expression that confusses me

do i use the chain rule on the (sinx+2cosx) because im unsure hows to differentiate the 2cosx
i know cosx goes to -sinx but what happens to the number at the front?

17. Dec 12, 2008

It's a constant, and d(c*f(x))/dx=c*d(f(x))/dx

As an example, the derivative of 2x is 2. Nothing changes about that when you do trig.

18. Dec 12, 2008

### HallsofIvy

Staff Emeritus
I am very confused as to what you are doing. Where are you getting these problems if not from a calculus book? And if you have a calculus book then surely it has the basic rules like (cf(x))'= c f'(x), (f+ g)'(x)= f'(x)+ g'(x), (fg)'(x)= f'(x)g(x)+ f(x)g'(x), (f/g)'(x)= (f'(x)g(x)- f(x)g'(x))/g2(x), and the chain rule. Those are what you need to learn.

19. Dec 12, 2008

### Fredrik

Staff Emeritus
You don't need the chain rule for 2 cos x. This is covered by the product rule: (fg)'(x)=f'(x)g(x)+f(x)g'(x). If f and g are defined by f(x)=2 and g(x)=cos x, the first term vanishes (f'(x)=0), so you're left with f(x)g'(x)=2 (-sin x).

Edit: Oops, I didn't realize that there was a second page of posts where this had already been answered.