I'm looking to figure out how to solve the following derivative..
(2e^(5x))(cos(3x) + sin(3x))
I know the answer to be..
(4e^(5x))(4cos(3x) + sin(3x))
But I can't figure out how to arrive at the answer. I've tried using the product rule, but I can't seem to get the same answer. I've also tried some logarithmic differentiation, but that doesn't seem to end up correct either. Could someone please explain to me how one goes about solving such a derivative.
2007-06-12 15:45:48 · 3 answers · asked by patrickkonsor 2 in Science & Mathematics ➔ Mathematics
2e^5x(cos(3x)+sin(3x)); product rule works out dude
=2e^5x(-3sin3x+3cos3x)+(cos 3x+sin 3x)(10e^5x)
=-6e^5xsin3x+6e^5xcos3x+10e^5xcos3x+10Ee^5xsin 3x
=4e^5xsin 3x+16e^5xcos 3x
=(4e^(5x))(4cos(3x) + sin(3x))
2007-06-12 16:06:41 · answer #1 · answered by harsha 1 · 0⤊ 0⤋
The product rule should work. d/dx(2e^5x) = 10e^5x. d/dx(cos[3x]+sin[3x])= -3 sin[3x] + 3 cos[3x]..... Perhaps the answer is wrong.
2007-06-12 22:58:01 · answer #2 · answered by cattbarf 7 · 0⤊ 0⤋
Both answers are right
use the product rule as above and then after some simplication it is equal to the above equation.
d(f(x)*g(x))/dx = d(f(x))/dx*g(x) + f(x)*d(g(x))/dx
Product rule
2007-06-12 23:10:24 · answer #3 · answered by Anonymous · 0⤊ 0⤋