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?

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

To find the derivative of the given function, y = e^2x * sin(3x), you correctly used the product rule. However, it seems that there was a small error in differentiating the exponential term.

Let's go through the steps of finding the correct derivative:

Step 1: Apply the product rule
The product rule states that the derivative of a product of two functions u(x) and v(x) is given by:
(d/dx) [u(x) * v(x)] = u'(x) * v(x) + u(x) * v'(x)

Step 2: Identify the functions u(x) and v(x)
In the given function, y = e^2x * sin(3x), we can consider u(x) = e^2x and v(x) = sin(3x).

Step 3: Find the derivative of u(x) and v(x)
To differentiate u(x) = e^2x, we use the chain rule:
u'(x) = 2e^2x
To differentiate v(x) = sin(3x), we use the chain rule as well:
v'(x) = 3cos(3x)

Step 4: Apply the product rule formula
Now, we can use the product rule formula:
(d/dx) [u(x) * v(x)] = u'(x) * v(x) + u(x) * v'(x)

Substituting the derivatives we found into the formula, we get:
(d/dx) [e^2x * sin(3x)] = (2e^2x * sin(3x)) + (e^2x * 3cos(3x))

Simplifying the expression, we find the correct derivative:
(d/dx) [e^2x * sin(3x)] = 2e^2x * sin(3x) + 3e^2x * cos(3x)

So, the actual answer is sin(3x) * (2e^2x) + cos^3x * (3e^2x). The 2 and 3 multiplying e^2x come from differentiating the exponential term using the chain rule.