Derivatives of inverse functions
Calculus Help

Question

Derivatives of inverse functions
Calculus Help

mathmusician · Answer

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mathmusician · Answer

I presume that I will need to find the inverse. Is that correct?

dumbcow · Answer

$$f'(x) = \frac{dy}{dx}$$
$$(f^{-1}(x))' = \frac{dx}{dy} = \frac{1}{dy/dx} = \frac{1}{f'(x)}$$

mathmusician · Answer

Sorry my computer is being weird for some reason,
After I do that where do I get the domain from?

zepdrix · Answer

Recall that the sine function achieves a maximum value of $\large\rm 1$,
this happens at an angle of $\rm \dfrac{\pi}{2}$.

Similarly, the sine function achieves a minimum value of $\large\rm -1$,
and this occurs at an angle of $\rm -\dfrac{\pi}{2}$.

In order for the inverse sine to be a `function`,
we have to restrict the domain so that it is one-to-one.|dw:1459790400453:dw|