Why derivative of $\frac{1}{\sin\:x}$ isn't same as $\sin\:x^{-1}$? According to $x^{-1}$ is same as $1/x$, therefore $\frac{1}{\sin\:x}$ should be same as $\sin\:x^{-1}$. Finding derivative of: 
$$f(x)=\frac{1}{\sin\:x}=\frac{0\cdot \sin x-1\cdot \left(-\cos x\right)}{\sin x^2}=f^{'}(x)=\frac{\cos x}{\sin x^{2}}$$.
$$f(x)=\sin\:x^{-1}=-\sin x^{-2}=f^{'}(x)=-\frac{1}{\sin x^{2}}$$
I got 2 different answers. Can u help me?
 A: Note that we usually write for clearness
$$(\sin x)^{-1}=\frac1{\sin x}\neq \sin (x^{-1})=\sin\left(\frac1x\right)\neq \sin^{-1}x=\arcsin x$$
then note that in your first derivation
$$f'(x)=\frac{0\cdot \sin x-1\cdot \left(\cos x\right)}{\sin^2 x}=-\frac{\cos x}{\sin^2 x}$$
in your second derivation we need to apply chain rule that is
$$\frac{d}{dx}[f(x)]^{-1}=-[f(x)]^{-2}f'(x)$$
A: You didn't apply the chain rule properly. Put
$$
f(x)=(\sin x)^{-1}
$$
Then 
$$
f'(x)=-(\sin x)^{-2}\times \frac{d}{dx}(\sin x)=-\frac{\cos x}{\sin^{2}x }
$$
In the first computation if
$$
f(x)=\frac{1}{\sin x}
$$
then using the quotient rule
$$
f'(x)=\frac{0\times\sin x-1\cos x}{\sin ^{2} x}
=\frac{-\cos x}{\sin ^2 x}$$
You need to properly distingush between $(\sin x)^2=\sin ^2 x$ and $\sin x^2=\sin (x^2) $ as well as $\sin x^{-1}=\sin (x^{-1})$ and $(\sin x)^{-1}$. Even though it may be clear to you which meaning you intend, use brackets to distinguish between them.
A: It looks like you are confusing the expressions $\sin(x^{-1})$ and $(\sin x)^{-1}$.  Basically, you are composing the sine function and the reciprocal function in both, but the order matters.
You are also misapplying the chain rule.
To find the derivative of the first:
$$
    \frac{d}{dx}(\sin x)^{-1} = (-1)(\sin x)^{-2} \frac{d}{dx} \sin x = - (\sin x)^{-2} \cos x = -\frac{\cos x}{\sin^2 x}
$$
And for the second:
$$
\frac{d}{dx}\sin(x^{-1}) = \cos(x^{-1}) \frac{d}{dx}(x^{-1}) = \cos(x^{-1})(-x^{-2})
$$
One last thing: you are writing an equals sign between a function and its derivative.  Only use an equals sign between quantities that are equal.  A function and its derivative are generally not.  It would be better to write
$$
    f(x) = \sin(x^{-1}) \color{blue}{\implies} f'(x) = -\cos(x^{-1})x^{-2}
$$
