Why derivative of $\frac{1}{\sin\:x}$ isn't same as $\sin\:x^{-1}$?

Question

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?

Answer 1

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.

Answer 2

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

$$\frac{d}{dx}[f(x)]^{-1}=-[f(x)]^{-2}f'(x)$$

Answer 3

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 $$ f(x) = \sin(x^{-1}) \color{blue}{\implies} f'(x) = -\cos(x^{-1})x^{-2} $$