Given: f(x) = e^2x, g(x) = sin 2x, and h(x) = 2x^2. Let p(x) = g[f(x)]. Then p'(x) =
2006-07-10 10:42:17 · 8 answers · asked by Olivia 4 in Science & Mathematics ➔ Mathematics
3 MINUTES!!!
OMG
2006-07-10 10:50:42 · update #1
3 MINUTES!!!
2006-07-10 10:51:58 · update #2
And I didn't even get it right
this is what I got...
4e2x cos(2e2x)
2006-07-10 10:59:32 · update #3
4e^2x cos(2e^2x)
2006-07-10 11:00:55 · update #4
g[f(x)] = sin[2(e^(2x))]
p(x) = sin[2e^(2x)]
p'(x) = cos[2e^(2x)] (2e^(2x))'
= cos[2e^(2x)](2e^(2x)) (2x)'
= cos[2e^(2x)](2e^(2x))(2)
= 4(e^(2x))cos[2e^(2x)]
2006-07-10 10:47:39 · answer #1 · answered by MsMath 7 · 12⤊ 1⤋
well, you can forget about h(x)...
p(x) = g[f(x)] = sin (2*e^2x)
let's see... the definition of d (sin x) = cos(x) dx, right?
The derivative of e^x = dx e^x
so...
so... p'(x) = cos (2*e^2x) * d(2e^2x) = 4e^2x cos(2e^2x)
hope that helps :-)
2006-07-10 17:52:46 · answer #2 · answered by bablunt 3 · 0⤊ 0⤋
are you supposed to get an actual number?
cuz like i got P(x) to equal e^4sinxcosx
but then maybe you could keep going i guess? i don't want to though.
i might be completely wrong but atleast i tried.
2006-07-10 17:50:43 · answer #3 · answered by -->JiLL<-- 3 · 0⤊ 0⤋
Okay, f( x) = e ^2x, g(x) = sin2x, and h(x) = 2x^2.
p(x) = g[f(x)].
so, p(x) = sin2( e^2x )
Then, p'(x) , would be cos2(e^2x) * 2*e^2x * 2.
( I think )
2006-07-10 19:29:14 · answer #4 · answered by M 1 · 0⤊ 0⤋
im rusty so i tried this.. i got to sin2(e^2x) ...where am i going wrong? how do you use h(x) ???
2006-07-10 18:09:16 · answer #5 · answered by Adam 2 · 0⤊ 0⤋
Oooh owwww, my brain!
2006-07-10 17:45:38 · answer #6 · answered by LadyRebecca 6 · 0⤊ 0⤋
Yesss I hate it too.
2006-07-10 17:45:59 · answer #7 · answered by Ricky J. 6 · 0⤊ 0⤋
4e^(2x)cose^(2x)
2006-07-10 17:46:46 · answer #8 · answered by lavi_or_lavinia 2 · 0⤊ 0⤋