One question I need help with

grandmastr

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Jun 27, 2005
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2
Y=sin(arcsin x)=x

Dy/Dx= Cos(arcsinx0)/square root of (1-x^2)=1

I need to show proof that the 2nd equation equals one.

Thanks for the help
 
Sorry, you're not making a lot of sense, so far.

You define "Y". Is that the same as the subsequently defined "y" in dy/dx?

You then have Cos(arcsinx0) in the numerator. What is that? Is the zero (0) extraneous? Did you mean cos(arcsin(x)) EVALUATED at x = 0 after finding the derivative?

Let's just try a wild guess:

cos(arcsin(x))

This needs a few things:
-- A Right Triangle,
-- Some Basic Trig Functions, and
-- The Pythagorean theorem

Draw a Right Triangle.
Label one of the Acute Angles, A
arcsin(x) gives an angle, the angle A, such that sin(A) = x.
Label two sides of your triangle so that sin(A) = x. That is, label the opposite leg, 'x', and label the hypotenuse '1'.
Calculate the adjacent leg, sqrt(1 - x^2)
Write an expression for cos(A). You should get sqrt(1-x^1)/1 or just sqrt(1-x^1).

Now, IF your original expression was supposed to be:

cos(arcsin(x))/sqrt(1-x^2)

That is pretty unfortunate, since that expression is equal to one (1), by the previous demonstration.

OK, another guess. The question is simply to find:

(d/dx)sin(arcsin(x)).

Using the chain rule, one gets, cos(arcsin(x))*(1/sqrt(1-x^2)), and we are left with proving this equal to unity. Hey, we did that already (See Above).
 
I just took the div. of the 1st equation and got the 2nd one.


Thanks for the help and I think i got the extra credit question right now.

So if I just follow each step of yours I should have this question right?
 
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