f(x) = [(e^x)+(e^-x)] / 2
2007-08-19 14:07:23 · 4 answers · asked by pandaeater9 2 in Science & Mathematics ➔ Mathematics
the derivative of e^x: (e^x)' = (x)'e^x or simply e^x.
Say you have e^3x, the derivative is (3x)'e^3x = 3e^3x. it is a property of differentiating e^x. Thus, the derivative of e^-x is merely -e^-x. Substituting, then:
f '(x) = (e^x - e^-x)/2.
there you go!:D i hope you'd understand
2007-08-19 14:35:45 · answer #1 · answered by mikael 3 · 0⤊ 2⤋
f (x) = (1/2) e^x + (1/2)e^(-x)
f `(x) = (1/2) e^x - (1/2) e^(-x)
f `(x) = (1/2) [ e^x - e^(- x) ]
2007-08-23 17:46:19 · answer #2 · answered by Como 7 · 0⤊ 0⤋
The derivative of e^x is e^x and the derivative of e^-x is -e^(-x). Therefore f'(x) = (e^x - e^(-x))/2.
f(x) is also known as cosh x and f'(x) is known as sinh x.
2007-08-19 21:12:44 · answer #3 · answered by Derek C 3 · 0⤊ 1⤋
I'm going to assume you know the following,
d/dx e^x = e^2
d/dx e^-x = -e^-x
f'(x)=(e^x-e^-x)/2
2007-08-19 21:11:20 · answer #4 · answered by de4th 4 · 0⤊ 1⤋