0
$\begingroup$

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.

  • 7
    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.2011-07-19
  • 0
    i have no where to start,also please tell me what necessary step i must take!2011-07-19
  • 0
    Thank you for expanding on your background. As for your question, [see here](http://en.wikipedia.org/wiki/Inverse_functions_and_differentiation).2011-07-19
  • 0
    @Arturo: Honest question, what is your opinion: In this situation, should I or should I not post a hint as I did below?2011-07-19
  • 0
    @Eric: Since you gave Victor the time to edit his question, I see no problem.2011-07-19
  • 0
    possible duplicate of my question: http://math.stackexchange.com/q/48752/78002011-07-19

2 Answers 2

5

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.

  • 0
    i think this is the best answer because it give me all necessary steps,thanks eric!2011-07-19
3

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.