Proving the Derivative of Inverse Sine without Trigonometry

[QuarkCharmer]

Homework Statement

Prove that:
\frac{d}{dx}arcsin(x) = \frac{1}{\sqrt{1-x^{2}}}

Homework Equations

The Attempt at a Solution

y=arcsin(x)
sin(y)=sin(arcsin(x))
sin(y)=x
\frac{d}{dx}(sin(y)=x)
cos(y)\frac{dy}{dx}=1
\frac{dy}{dx}=\frac{1}{cos(y)}
Using a triangle:
cos(y)=\sqrt{1-x^2}
\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}
\frac{d}{dx}arcsin(x) = \frac{1}{\sqrt{1-x^{2}}}

We were shown these in class before starting trig-sub and all that, and the professor said that there are proofs by differentiation. I don't know how to explain the steps where I determine that the cosine of the angle y is equal to sqrt(1-x^2) without drawing out a triangle. Is there some way? Other than that I think I have it figured out and I can do the same for the other functions.

 

[lineintegral1]

Well, certainly,

x=sin(y)

In addition,

sin^2y+cos^2y=1

So it certainly follows that,

cos(y)=\sqrt{1-sin^2y}

You can see that it follows naturally (by substitution) that,

cos(y)=\sqrt{1-x^2}

And we're done!

 

[QuarkCharmer]

Oh haha. That really should have occurred to me considering that I used the pyth. theorem to find cos(y).

Thanks

 

[HallsofIvy]

Since you refer to "Using a triangle", you can also do it this (equivalent) way:
imagine a right triangle triangle having "opposite side" of length x and "hypotenuse" of length 1, so that sin(y)= x/1= x. Then, by the Pythgorean theorem, the "near side" has length \sqrt{1- x^2}. From that, cos(y)= \sqrt{1- x^2}.