Find (d(cos x))/(d(sin x))
Find $\dfrac{d(\cos x)}{d(\sin x)}$.
Explanation of Method
A derivative taken with respect to another function.
Step 1: Interpret the symbol as a ratio of differentials
Both $\cos x$ and $\sin x$ are functions of the same variable $x$, so each has a differential; the quotient asks how one changes per unit change of the other: $\frac{d(\cos x)}{d(\sin x)} = \frac{(\cos x)'\,dx}{(\sin x)'\,dx} = \frac{-\sin x}{\cos x}$.
The common factor $dx$ cancels, which is exactly why the ratio is well defined.
Step 2: Simplify
\[ \frac{d(\cos x)}{d(\sin x)} = -\tan x. \]
Step 3: Verification
Substituting $u = \sin x$ makes the question an ordinary derivative. On an interval where $\cos x \gt 0$ we have $\cos x = \sqrt{1-u^{2}}$, so
\[\frac{d(\cos x)}{d(\sin x)} = \frac{d}{du}\sqrt{1-u^{2}} = \frac{-u}{\sqrt{1-u^{2}}} = \frac{-\sin x}{\cos x} = -\tan x \;\;\checkmark\]
\[x = \frac{\pi}{4}:\quad -\tan\frac{\pi}{4} = -1, \text{ matching } \frac{-\sin(\pi/4)}{\cos(\pi/4)} = -1 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,\dfrac{d(\cos x)}{d(\sin x)} = -\tan x, \qquad \cos x \neq 0\,}\]
Remark
Remark. This is the parametric derivative formula $\dfrac{dy}{dx} = \dfrac{y'_{t}}{x'_{t}}$ with $t = x$, $x_{\text{new}} = \sin t$, $y_{\text{new}} = \cos t$ — that is, the slope along the unit circle traced by $(\sin t,\cos t)$. Compare Problem 191, where the same circle is parametrised the other way round and the answer comes out as $-\cot t$.