Y = e^x: derivative is e^x?? pre-calc

if f(x) = e^x then f ' (x) = e^x then how come ...

g(x) = e^(2x+4)
g ' (x) = 2 e^(2x+4) .... Where did the 2 in the front come from????


can some1 explain to me plxxx! ty!!! with another eg!

You must differentiate the power as well. Hidden in the fact that if f(x) = e^x then f'(x) = e^x as well. What is left out is that

f'(x)= d(x)/dx * d(e(x))/dx = 1 * e^x = e^x.

That 1 came from differentiating the x that is the power.

So you have to differentiate not only e^(2x + 4) to get e^(2x + 4), you have to differentiate 2x + 4

d(2x + 4)/dx = 2, and that is the mystery of where the 2 came from.

When trying to find a derivative, you have to derive any other function that are in the global function and multiply it to the total (by the chain rule).

In your example, g(x) = e^(2x+4). Keep in mind that the derivative of e^x is e^x. So your derivative of e^(2x+4) should give e^(2x+4).

But your x in e^x is replaced by the function f(x) = 2x+4.
So you have g(x) = e^f(x). (if you simplify g(x) = e^(2x+4).

g'(x) = e^f(x) . f'(x), because you have to derivate any other function in your global function and multiply it to the total. f'(x) = 2.

So you get g'(x) = 2 e^(2x+4).
:)

Yes, chain rule gives you the correct derivative.
When in doubt, look at the facts. Here are some values for f and an approximation for f' based on h = 0.001

___f(x)= e^(2x+4)____ 2e^(2x+4)___ (f(x+h)-f(x))/h
0.5000 148.4131591 296.8263182 296.86
0.5001 148.4428447 296.8856894 -
1.5000 1096.633158 2193.266317 2,193.49
1.5001 1096.852507 2193.705014 -
2.5000 8103.083928 16206.16786 16,207.79
2.5001 8104.704706 16209.40941 -
3.5000 59874.14172 119748.2834 119,760.26
3.5001 59886.11774 119772.2355 -
4.5000 442413.392 884826.784 884,915.27
4.5001 442501.8835 885003.7671 -
5.5000 3269017.372 6538034.745 6,538,688.59
5.5001 3269671.241 6539342.483 -

f(x) = e^(x) d/dx 1

note the derivative of x is 1 so

f'(x) = e^(x)

g(x) = e^(2x + 4)
g'(x) = e^(2x + 4) d/dx 2

note we solve the derivative of 2x + 4 which is 2

so

g'(x) = 2e^(2x + 4) answer//

from chain rule

dy/dx = dy/du[du/dx]

y = e^[2x+4] u = 2x+4

dy/dx = de^u]/dud[2x+4]/dx = e^u [2] = 2e^[2x+4]

Chain rule.
d/dx(e^u) = du/dx * e^u
d/dx(e^(2x + 4)) =
d/dx(2x + 4) * e^(2x + 4) =
2e^(2x + 4)

(e^u)' = (e^u) * du {chain rule}