Derivative of a function w.r.t. another function. 
How is this? I'm getting $(-\tan(x))$. Here's my attempt:
Let $u=\sin(x)$ and $v=\cos(x)$. Then, the derivative we seek is,

$$\frac{\mathrm dv}{\mathrm du}$$

Using chain rule, we have,
$$\frac{\mathrm dv}{\mathrm du}=\frac{\mathrm dv}{\mathrm dx}\cdot \frac{\mathrm dx}{\mathrm du}=(-\sin(x))\cdot\frac{1}{\cos(x)}=(-\tan(x))$$

I can't find my flaw. Please help.

Here's the W|A link (shortened by Bit.ly) if anyone wants to verify.
 A: To do the calculation a more plausible way,
$$ \frac{d}{d\sin{x}}\cos{x} = \frac{d}{d\sin{x}}\sqrt{1-\sin^2{x}} = \frac{d}{dy}\sqrt{1-y^2}, $$
by writing $y=\sin{x}$ (mere replacement of symbols, doesn't mean anything more)
$$ \frac{d}{dy}\sqrt{1-y^2} = -\frac{y}{\sqrt{1-y^2}} = -\frac{\sin{x}}{\cos{x}} = -\tan{x}, $$
so your answer is correct.
My guess is that W|A is misinterpreting your input: for example, this doesn't do what you would expect either. It's probably just interpreting $d$ as a constant, so it thinks it is just doing fraction arithmetic.
A: To find $\frac{dx}{du}$ we see $x=\arcsin u$. Then,
$$\frac{dx}{du}=\frac{1}{\sqrt{1-u^2}}=\frac{1}{\sqrt{1-\sin^2 x}}=\frac{1}{\cos x}$$
so that step isn't wrong like Nitin said. I don't see why your solution is wrong, I'd love to see insight anybody else has.
As some others have said, W|A must just be canceling the $d$'s. Your answer is A-okay!!
A: WA interprets the $d$ as a constant, so they cancel out. The resulting fraction, $\frac{\cos x}{\sin x}=\cot x$. 
