# derivative of the function y=e^x

Printable View

• December 16th 2010, 09:52 AM
advancedfunctions2010
derivative of the function y=e^x
f(x)= (e^x +1)^2
I am not sure how to find the derivative of this one!
Do I (e^x +1)(e^x +1)?
Thanks in advance!
• December 16th 2010, 09:55 AM
snowtea
Use the chain rule.

du^2/dx = 2u du/dx

Where u = e^x + 1
• December 16th 2010, 11:29 AM
janvdl
Quote:

Originally Posted by advancedfunctions2010
f(x)= (e^x +1)^2
I am not sure how to find the derivative of this one!
Do I (e^x +1)(e^x +1)?
Thanks in advance!

$f(x) = (e^x + 1)(e^x + 1)$

$f'(x) = (e^x)(e^x + 1) + (e^x + 1)(e^x) = 2e^{2x} + 2e^x$

Or I suppose you could have said:

$f(x) = e^{2x} + 2e^x + 1$

$f'(x) = 2e^{2x} + 2e^x + 0 = 2e^{2x} + 2e^x$
• December 16th 2010, 11:34 AM
HallsofIvy
Or simply use the chain rule as snowtea suggested:
$f'(x)= 2(e^x+ 1)(e^x+1)'= 2(e^x+ 1)(e^x)= 2e^{2x}+ 2e^x$