SOLUTION: Please Find the Derivatives of the following and simplify: y=5e^3x y=1/e^x y=e^3 y=e^x2e^2x Thanks!

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Question 59236: Please Find the Derivatives of the following and simplify:
y=5e^3x

y=1/e^x

y=e^3

y=e^x2e^2x

Thanks!
Found 2 solutions by venugopalramana, funmath:
Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
Please Find the Derivatives of the following and simplify:
y=5e^3x
DY/DX = 5*3E^3X=15E^3X
y=1/e^x=E^(-X)
DY/DX = -1E^(-X)=-1/E^X
y=e^3 =CONSTANT
DY/DX =0
y=e^x2e^2x
DO YOU MEAN
Y=[E^X][2E^2X]..IF SO
DY/DX = (E^X)[2*2E^(2X)+2E^2X]= 6(E^X)(E^2X)
----IF YOU MEANT [E^(X^2)][E^2X]..THEN
DY/DX = [E^(X^2)]2E^(2X)+[E^2X]2XE^(X^2)=2[E^(X^2)][E^2X][1+X)
Thanks!

Answer by funmath(2933) About Me  (Show Source):
You can put this solution on YOUR website!
Find the Derivatives of the following and simplify:
d/dx[cf(x)]=cf'(x)
d/dx[e^x]=e^x
d/dx[c]=0
The chain rule: d/dx[f(g(x)]=f'(g(x))g'(x)
:
y=5e%5E%283x%29 Use the chain rule:
dy%2Fdx=5e%5E%283x%29%2A%283%29
dy%2Fdx=15e%5E%283x%29
:
y=1%2Fe%5Ex ---->> y=e%5E%28-x%29 Use the chain rule again:
dy%2Fdx=e%5E%28-x%29%2A%28-1%29
dy%2Fdx=-e%5E%28-x%29
dy%2Fdx=-1%2Fe%5Ex
:
y=e%5E3
There is no x, this is a constant even though it doesn't look like one. e%5E3=20.08553692. The derivative of a constant is 0.
dy%2Fdx=0
:
I'm not sure if I'm interpretting you right on this one, let me know if there's a misunderstanding!!!
y=e%5Ex%2A2e%5E2x-------->> y=2e%5E%28x%2B2x%29--->> y=2e%5E%283x%29 Use chain rule:
dy%2Fdx=2e%5E%283x%29%2A%283%29
dy%2Fdx=6e%5E%283x%29
:
Happy Calculating!!!