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

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In summary, the conversation discusses the differentiation of e^-x and e^(3x+4) and the use of the chain rule and product rule in solving these problems. The experts advise going back to basics and understanding the rules and concepts before attempting these problems.
  • #1
matttan
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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) ?

:confused:
 
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  • #2
You mean d(e^-x)/dx

Do you know what is df[g(x)]/dx in general?
 
  • #3
epenguin said:
You mean d(e^-x)/dx

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

I will use other notation:smile:

If f(x)=e^-x then do f'(x)=e^-x ?
 
  • #4
epenguin said:
Do you know what is df[g(x)]/dx in general?

I do not know, please enlight me.
 
  • #5
matttan said:
I will use other notation:smile:

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

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

(And apart from calculation it is possible to visualise it.)
 
  • #6
epenguin said:
In another notation do you know what [f(g(x))]' is ?

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

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
matttan said:
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) ?

:confused:

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
matttan said:
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?

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
matttan said:
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) ?

:confused:

For question #1, e-x=[tex]\frac{1}{e^x}[/tex]
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)
 
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  • #10
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
epenguin said:
OK - what is f'(-x) in general??

Sorry, I meant what is [f(-x)]' in general?
 
  • #12
d/dx of e^-x = -e^-x

if all else fails or you want further explanation 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)

is that the answer your looking for?
 
  • #13
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
matttan said:
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) ?

:confused:
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
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.
 
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  • #16
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 I am unsure hows to differentiate the 2cosx
i know cosx goes to -sinx but what happens to the number at the front?
 
  • #17
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
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
matt_crouch said:
do i use the chain rule on the (sinx+2cosx) because I am unsure hows to differentiate the 2cosx
i know cosx goes to -sinx but what happens to the number at the front?
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.
 

1. What is the derivative of e^-x?

The derivative of e^-x is -e^-x.

2. How do you prove that d/dx e^-x = e^-x?

To prove that d/dx e^-x = e^-x, we can use the definition of the derivative, which states that the derivative of a function f at a point x is equal to the limit as h approaches 0 of (f(x+h) - f(x))/h. By plugging in e^-x for f(x), we get the limit as h approaches 0 of (e^-(x+h) - e^-x)/h. We can then simplify this using the properties of exponents and algebra to get the simplified form of -e^-x. This shows that the derivative of e^-x is indeed -e^-x.

3. Can you provide an example of using the derivative of e^-x in a real-world application?

The derivative of e^-x is used to model many real-world phenomena, such as radioactive decay, population growth, and cooling of objects. For example, in radioactive decay, the rate at which a radioactive substance decays is proportional to the amount of substance present, which can be represented by the function e^-x. The derivative of this function, -e^-x, represents the rate of decay at any given time.

4. How does the derivative of e^-x relate to the graph of e^-x?

The derivative of e^-x, -e^-x, is the slope of the tangent line at any point on the graph of e^-x. This means that the slope of the tangent line is always negative and decreases as x increases. The graph of -e^-x is also a decreasing curve, which is reflected in the decreasing slope of the tangent line.

5. Is the derivative of e^-x always negative?

Yes, the derivative of e^-x, -e^-x, is always negative. This is because e^-x is a decreasing function, meaning that as x increases, e^-x decreases. Therefore, the slope of the tangent line, represented by -e^-x, is always negative.

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