Calculus Problem-first derivative of logs

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Belby

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If f(x)=e^3x+6 find the first derivative of f(x). I have the following but it is not right according to my computer math program:
f(x)=e(3x+6)^1/2
f ' (x)=e^(3x+6)^1/2 *(3x+6)^1/2'
f ' (x)=e^(3x+6) * 1/2(3x+6)^-1/2 * 3
f ' (x)=3/2 e^3x+6/^3x+6
 
You want the derivative of \(\displaystyle e^{3x+6}\)?.

Just apply the chain rule and get \(\displaystyle 3e^{3x+6}\)
 
Belby said:
If f(x)=e^3x+6 find the first derivative of f(x). I have the following but it is not right according to my computer math program:
f(x)=e(3x+6)^1/2 <<<< Where did this ^1/2 come from?
f ' (x)=e^(3x+6)^1/2 *(3x+6)^1/2'
f ' (x)=e^(3x+6) * 1/2(3x+6)^-1/2 * 3
f ' (x)=3/2 e^3x+6/^3x+6
Click to expand...
 
the 1/2 came when I took 3x+6 out of the square root. Is this not correct?
 
Belby said:
the 1/2 came when I took 3x+6 out of the square root. Is this not correct?
Click to expand...
There was no square root in your original equation.

f(x)=e^3x+6
 
Belby said:
yes, 3x+6 is under the radical sign
Click to expand...

So you should have written

y = e^[(3x+6)^(1/2)] ............. when you post incorrect problem - you waste both your time and our time.

Now use chain rule:

You have

y = e^[g(x)]

where

g(x) = (3x+6)^(1/2)

g'(x) = (1/2) * 3 * (3x+6)^(-1/2)

then

y' = g'(x) * e^[g(x)]

Now finish it ....
 
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