The Maclaurin series for the derivative of 3x^2 cos(5x) is -3x^3 + x^5 - x^7 + ...
To find the derivative of the Maclaurin series for 3x^2 cos(5x), we can use the following identity:
d/dx [cos(x)] = -sin(x)
This identity tells us that the derivative of cos(x) is -sin(x). We can use this identity to find the derivative of the Maclaurin series for 3x^2 cos(5x) as follows:
f'(x) = d/dx [3x^2 cos(5x)] = 3x^2 d/dx [cos(5x)] = -3x^2 sin(5x)
The Maclaurin series for sin(x) is:
sin(x) = x - x^3/3! + x^5/5! - x^7/7! + ...
Substituting this into the expression for f'(x), we get:
f'(x) = -3x^2 (x - x^3/3! + x^5/5! - x^7/7! + ...) = -3x^3 + x^5 - x^7 + ...
Therefore, the Maclaurin series for the derivative of 3x^2 cos(5x) is -3x^3 + x^5 - x^7 + ...
Question
2. Find the Maclaurin series for f(c) = 3x^2 cos(5x).
3. Find the Maclaurin series for the derivative of 3x^2 cos(5x). (Hint: Use the answer above and find the derivative of the series for 3x^2 cos(5x^2).)
Find the Maclaurin series for 3x^2 cos(5x) dx.