Calculus (3rd Edition)

Recall the product rule: $(uv)'=u'v+uv'$ Recall that $(e^x)'=e^x$ Recall that $\cos(-x)=\cos(x)$. Recall that $\sin(-x)=-\sin(x)$. Since we have $$y= \cos (e^\theta )$$ then the first derivative $\frac{dy}{d\theta}$, using the chain rule, is given by \begin{align*} \frac{dy}{d\theta}&= -\sin(e^\theta)(e^\theta)'\\ &= -e^\theta \sin(e^\theta). \end{align*} The second derivative $\frac{d^2y}{d\theta}$, using the product rule, is given by \begin{align*}\frac{d^2y}{d\theta^2}&=-(e^\theta)'\sin(e^\theta)-(e^\theta)(\sin(e^\theta))' \\ &=-e^\theta\sin(e^\theta)-e^{2\theta} \cos(e^\theta)\\ &=-e^\theta(\sin(e^\theta)+e^{\theta} \cos(e^\theta)). \end{align*} The third derivative $\frac{d^3y}{d\theta}$, using the product rule, is given by \begin{align*}\frac{d^3y}{d\theta}&=-(e^\theta)' (\sin(e^\theta)+e^{\theta} \cos(e^\theta))-e^\theta(\sin(e^\theta)+e^{\theta} \cos(e^\theta))' \\ &=-e^\theta (\sin(e^\theta)+e^{\theta} \cos(e^\theta))-e^\theta(e^\theta\cos(e^\theta)+e^\theta\cos(e^\theta)-e^{2\theta} \sin(e^\theta)) \\ &= e^{3 \theta} \sin \left(e^{\theta}\right)-3 e^{2 \theta} \cos \left(e^{\theta}\right)-e^{\theta} \sin \left(e^{\theta}\right). \end{align*}