## Calculus (3rd Edition)

$$\frac{d y}{d x}= \frac{-\sin \theta}{2\tan\theta \sec^2\theta}$$ and at $\theta=\pi/4$, we have $$\frac{d y}{d x}= -\frac{\sqrt 2}{8} .$$
Since $x=\tan^2\theta$ and $y=\cos\theta$, then we have $$\frac{d y}{d x}=\frac{y^{\prime}(t)}{x^{\prime}(t)}=\frac{-\sin \theta}{2\tan\theta \sec^2\theta}$$ and at $\theta=\pi/4$, we have: $$\frac{d y}{d x}=\frac{-\frac{\sqrt{2}}{2}}{2\times1\times(\sqrt{2})^2}$$ Which simplifies to: $$\frac{d y}{d x}= -\frac{\sqrt 2}{8}.$$