Does Differentiating e^(-x) and e^(3x+4) Follow Standard Rules?

[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]

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]

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]

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]

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)

 

[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)

 

[epenguin]

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 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?

 

[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]

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)= e^g and, apparently, you know that d e^g/dg= e^g. What is dg/dx= d(-x)/dx?
Multiply those together.

 

[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 e^x is always positive and always increases with x. I.e. d(e^x)/dx is always positive.

So e^-x which is 1/e^x, 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.

 

[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 I am unsure hows to differentiate the 2cosx
i know cosx goes to -sinx but what happens to the number at the front?

 

[Dr. Lady]

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]

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.

 

[Fredrik]

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.