differentiating sinx in relation to time

Question

So I am trying to figure this out, and using chain rule,I get 1(sinx)^0 x cos dx/dt, but somehow the answer is cos x dx/dt. I guess I can't figure out where the extra x is coming from because while I get that it is cos x normally, in this case doesn't the x become dx/dt? If anyone can explain this to me, I would appreciate it greatly.

Answer 1

Recall the chain rule: $\frac{d}{dx}f(g(x)) = f'(g(x)) g'(x)$.

I presume that you are trying to differentiate $\sin(x(t))$ with respect to time. Let $f=\sin$ and let $g=x$, and let $x=t$. Then use the chain rule as stated above.

$\frac{d}{dt}\sin(x(t))=\sin'(x(t))\cdot x'(t)=\cos(x(t)) \cdot x'(t)$, which is the stated answer.