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Find the derivative of the function $\sin^{-1}x$ assuming $\frac{d}{dx} \sin x = \cos x$ has been found. Justify your answer.

It is just a book that I bought without answers and I am a high school student, I am just wondering how to finish the proof (I need a complete answer, but a hint is ok).

Thanks in advance.

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Victor: You've been on this site for 20 days; you've posted 20 questions. Surely you should know by now that many of us consider it extremely rude for you to post in the imperative ("Find", "Justify", "Prove") as if you were giving us an assignment or homework. Please rephrase the question as a question, indicating where you are having trouble and why. – Arturo Magidin Jul 19 '11 at 16:48
i have no where to start,also please tell me what necessary step i must take! – Victor Jul 19 '11 at 17:00
Thank you for expanding on your background. As for your question, see here. – t.b. Jul 19 '11 at 17:01
@Arturo: Honest question, what is your opinion: In this situation, should I or should I not post a hint as I did below? – Eric Naslund Jul 19 '11 at 17:01
@Eric: Since you gave Victor the time to edit his question, I see no problem. – Arturo Magidin Jul 19 '11 at 17:06
up vote 5 down vote accepted

Hint: For $x\in(-1,1)$, $$\sin\left(\sin^{-1}(x)\right)=x.$$ Taking the derivative of both sides we get $$\cos\left(\sin^{-1}(x)\right)\cdot \left(\frac{d}{dx} \sin^{-1}(x)\right)=1.$$

Now, all that is left is to find $\cos\left(\sin^{-1}(x)\right)$. For this, I suggest drawing a right angle triangle with hypotenuse $1$, and side length $x$ so you can see exactly how $\sin^{-1}$ and $\cos$ interact.

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i think this is the best answer because it give me all necessary steps,thanks eric! – Victor Jul 19 '11 at 17:09

Here's one way of looking at it: Let $y = \arcsin x$ so $x = \sin y$. Then $\frac{dx}{dy} = \cos y$, so $\frac{dy}{dx} = \frac{1}{\cos y}$. Since $\sin^2 y + \cos^2 y = 1$, you can change $\frac{1}{\cos y}$ to $\frac{1}{\sqrt{1 - \sin^2 y}} = \frac{1}{\sqrt{1-x^2}}$.

Two questions arise: why is there no "$\pm$" in front of the radical? and is the reciprocal of $\frac{dx}{dy}$ really $\frac{dy}{dx}$ even though $dy$ and $dx$ aren't actually numbers?

The answer to the first question comes from the fact that if $y = \arcsin x$ then $y$ is between $-\pi/2$ and $\pi/2$, so that $\cos y$ is nonnegative (either positive or zero).

The answer to the second question is "yes" because of the chain rule.

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