find the derivative of g (theta)=sine squared (3 theta)
g = sin^2(3T)
dg/dT = 2 sin(3T) (3) cos(3T)
= 6 sin (3T) cos(3T)
= 3sin(6T)
yes, cool :)
To find the derivative of g(theta) = sin^2(3theta), we can apply the chain rule.
The chain rule states that if we have a function of the form f(g(theta)), then the derivative is given by:
f'(g(theta)) * g'(theta)
In this case, f(u) = u^2, and g(theta) = sin(3theta).
First, let's find the derivative of g(theta), which is g'(theta):
g'(theta) = d/dtheta [sin(3theta)]
To find the derivative of sin(3theta), we can use the chain rule again.
Let's define u = 3theta. Then, we have:
g'(theta) = d/du [sin(u)] * du/dtheta
Now, the derivative of sin(u) with respect to u is cos(u):
g'(theta) = cos(u) * du/dtheta
To find du/dtheta, we need to find the derivative of u = 3theta with respect to theta:
du/dtheta = d/dtheta [3theta]
This derivative is straightforward, as we are differentiating with respect to theta:
du/dtheta = 3
Substituting these back into g'(theta), we have:
g'(theta) = cos(u) * du/dtheta
= cos(3theta) * 3
Now, we can find the derivative of g(theta) using the chain rule:
g'(theta) = f'(g(theta)) * g'(theta)
= 2 * sin(3theta) * cos(3theta) * 3
= 6 sin(3theta) * cos(3theta)
Therefore, the derivative of g(theta) = sin^2(3theta) is:
g'(theta) = 6 sin(3theta) * cos(3theta)