Given $y(x)$ and $x=\sin\theta$, how would you differentiate the following equation w.r.t $x$
$$\frac{dy}{dx} = (\sec\theta) \frac{dy}{d\theta}$$
I've tried to differentiate using the product rule but one of the terms is:
$$\sec\theta\frac{d}{dx}\left\{\frac{dy}{d\theta}\right\}$$
I'm not sure how to deal with this term to obtain an equation in terms of $y$ and $\theta$. Thanks.
$$\frac{dy}{dx} = (\sec\theta) \frac{dy}{d\theta}$$ do you mean this ? $$\frac{d^2y}{dx^2} =\frac d {dx}\left( (\sec\theta) \frac{dy}{d\theta} \right )$$ $$\frac{d^2y}{dx^2} =\frac d {d \theta}\left( (\sec\theta) \frac{dy}{d\theta} \right ) \frac {d \theta}{dx}$$ $$\frac{d^2y}{dx^2} =\left( (\sec\theta) \frac{d^2y}{d\theta^2} + \frac{dy}{d\theta}\frac d {d \theta} \sec \theta\right ) \frac {d \theta}{dx}$$ $$\frac{d^2y}{dx^2} =\left( (\sec\theta) \frac{d^2y}{d\theta^2} + \frac{dy}{d\theta}\frac d {d \theta} \sec \theta\right ) \sec \theta$$
And $$\frac d {d \theta} \sec \theta=\frac {\sin \theta}{\cos^2 \theta}$$ So that we finally have $$\frac{d^2y}{dx^2} =\left( \frac{d^2y} {d\theta^2} + \tan \theta\frac{dy}{d\theta} \right ) \sec^2 \theta$$