I was going through a textbook of mine and I noticed that in the proof of derivative of $y = \sin^{-1}$,
i.e. proof that $$\frac{dy}{dx} \sin^{-1}(x) = \frac{1}{\sqrt{1-x^2}}$$
there's a point where they say $$\frac{1}{\cos y} = \frac{1}{\sqrt{1-x^2}}$$
and I'm not sure how this makes sense or works out. I've looked for proofs, tried implicit differentiation, and tried graphing it, but couldn't find anything. Can someone please explain?
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You actually can prove this using implicit differentiation. In fact, you should do it to demonstrate why this seemingly confusing fact is true. Here are some steps:
1) set x = sin(y) and apply the derivative in respect to x on both sides and use the chain rule to evaluate the right side. You should find that 1/cos(y) = dy/dx
2) Now, use the Pythagorean identity in respect to y and solve for cos(y). Remember what we found dy/dx equals above. Plug in what you found cos(y) is based on the Pythagorean identity. What did we, at the very beginning, say that sin(y) was equal to prior to differentiating both sides? Now you should see why we say that is the derivative of sin.
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