합성함수의 미분법

At time $t$, the particle is at the point $\big(x(t),y(t)\big)$, with $x(t)=\cos t$ and $y(t)=\sin t$. Over time, the particle traces out a curve; let's call that curve $y=f(x)$. Then $y(t) = f\big(x(t)\big)$, so the slope of the curve at the point $\big(x(t),y(t)\big)$ is $f'\big(x(t)\big)$. You are to determine the values of $t$ for which $f'\big(x(t)\big)=-1$.

By the chain rule

$$\begin{aligned} y'(t) = f'\big(x(t)\big) \cdot x'(t) \end{aligned}$$

Substituting in $x(t)=\cos t$ and $y(t)=\sin t$ gives

$$\begin{aligned} \cos t = f'\big(x(t)\big) \cdot \big(-\sin t\big) \end{aligned}$$

so that

$$\begin{aligned} f'\big(x(t)\big) = -\frac{\cos t}{\sin t} \end{aligned}$$

is $-1$ precisely when $\sin t = \cos t$. This happens whenever $t = \frac{\pi}{4}$.

Remark: the path traced by the particle is a semicircle. You can think about the point on the unit circle with angle t, or you can notice that $x^2 + y^2 = \sin^2t + \cos^2t = 1$.