Derivative of Composite Trigonometric Function
To find the derivative of the function \( y = \cos(e^{3\theta^2}) \), we will apply the chain rule, which says that the derivative of a composite function is the derivative of the outer function times the derivative of the inner function.
Let \( u = e^{3\theta^2} \), so our function becomes \( y = \cos(u) \).
Taking the derivative with respect to \(\theta\), we get:
\( \frac{dy}{du} = -\sin(u) \) (derivative of the outer function, \( \cos(u) \) with respect to \( u \))
Now, we find \( \frac{du}{d\theta} \):
\( u = e^{3\theta^2} \) implies \( \frac{du}{d\theta} = \frac{d}{d\theta} e^{3\theta^2} \)
To find this, we use the chain rule again:
Let \( v = 3\theta^2 \), then \( u = e^v \)
Take the derivative of \( v \) with respect to \( \theta \):
\( \frac{dv}{d\theta} = 6\theta \)
Now, take the derivative of \( u \) with respect to \( v \):
\( \frac{du}{dv} = e^v \)
So the derivative of \( u \) with respect to \( \theta \) is:
\( \frac{du}{d\theta} = \frac{du}{dv} \cdot \frac{dv}{d\theta} = e^v \cdot 6\theta = e^{3\theta^2} \cdot 6\theta \)
Now, we use the chain rule to find the derivative of \( y \) with respect to \( \theta \):
\( \frac{dy}{d\theta} = \frac{dy}{du} \cdot \frac{du}{d\theta} = -\sin(u) \cdot e^{3\theta^2} \cdot 6\theta \)
Substitute \( u \) back in to get:
\( \frac{dy}{d\theta} = -\sin(e^{3\theta^2}) \cdot e^{3\theta^2} \cdot 6\theta \)
So the answer is:
\[ \frac{dy}{d\theta} = -6\theta e^{3\theta^2} \sin(e^{3\theta^2}) \]
