Question - Derivative of Composite Trigonometric Function
Solution:
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:\[ \frac{dy}{d\theta} = -6\theta e^{3\theta^2} \sin(e^{3\theta^2}) \]
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