Question

Найти y′(0), если y(x)=e^2x sin3x

Answer 1

у"=2е^2x*sin3x+e^2x*3cos3x=e^2x(2sin3x+3cos3x)

y"(0)=e в 0 степ.(2sin0+3cos0)=1*(0+3*1)=3

Answer 2

To find y'(0), we need to find the derivative of the function y(x) with respect to x and then evaluate it at x = 0.

Given the function y(x) = e^(2x) * sin(3x), we can find y'(x) by applying the product rule and chain rule.

Let's find the derivative step by step:

1. Apply the product rule: (uv)' = u'v + uv'

   Let u = e^(2x) and v = sin(3x). Then, u' = d/dx(e^(2x)) and v' = d/dx(sin(3x)).

2. Find u': u = e^(2x) Applying the chain rule, we have: u' = d/dx(e^(2x)) = 2e^(2x)

3. Find v': v = sin(3x) Applying the chain rule, we have: v' = d/dx(sin(3x)) = 3cos(3x)

4. Apply the product rule: (uv)' = u'v + uv' (e^(2x) * sin(3x))' = (2e^(2x) * sin(3x)) + (e^(2x) * 3cos(3x))

Now, let's evaluate y'(0):

x = 0 (2e^(2(0)) * sin(3(0))) + (e^(2(0)) * 3cos(3(0))) (2 * 1 * 0) + (1 * 3 * 1) 0 + 3 3

Therefore, y'(0) = 3.