Question
y = e^2xsin3x Find the derivative.
So I used the product rule and I got sin3x(e^2x)+cos^3x(e^2x). But the actual answer is sin3x(2e^2x)+cos^3x(3e^2x) Where the the 2 and 3 before the e come from?
So I used the product rule and I got sin3x(e^2x)+cos^3x(e^2x). But the actual answer is sin3x(2e^2x)+cos^3x(3e^2x) Where the the 2 and 3 before the e come from?
Answers
Damon
I guess you mean
[e^(2x)]sin 3x
sin 3x [2 e^(2x)] + 3 cos (3x) e^(2x)
hey same mistake as earlier
d/dx (e^u) = du/dx * e^u
and
d/dx (sin u) = du/dx cos u
if u = 2x, then du/dx = 2
if u = 3x, then du/dx = 3
[e^(2x)]sin 3x
sin 3x [2 e^(2x)] + 3 cos (3x) e^(2x)
hey same mistake as earlier
d/dx (e^u) = du/dx * e^u
and
d/dx (sin u) = du/dx cos u
if u = 2x, then du/dx = 2
if u = 3x, then du/dx = 3