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

#### 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) ? #### epenguin

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You mean d(e^-x)/dx

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

#### matttan

You mean d(e^-x)/dx

Do you know what is df[g(x)]/dx in general?
I will use other notation If f(x)=e^-x then do f'(x)=e^-x ?

#### matttan

Do you know what is df[g(x)]/dx in general?
I do not know, please enlight me.

#### epenguin

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I will use other notation 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.)

#### matttan

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?

#### aLiase

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) ? 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

#### epenguin

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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??

#### symbolipoint

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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) ? 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)

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#### symbolipoint

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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)

#### epenguin

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

#### 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)

#### 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.)

#### HallsofIvy

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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) ? 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.

#### epenguin

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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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#### 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?

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.

#### HallsofIvy

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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.

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