Expert: Paul Klarreich - 10/15/2006

Question
Hi,

Could you advise me how to get started with the following question.

y= INV Sin (x/5)

Should i change Sin from inverse by saying 1/[Sin (x/5)]

Please advise as I am unsure.

Many thanks.
Annie

Answer
Annie Asks in Category Calculus ...
 
Subject:  Differentiating Trigonometric Equations
 
Question:  Hi,

Could you advise me how to get started with the following question.

y= INV Sin (x/5)

Should i change Sin from inverse by saying 1/[Sin (x/5)]

Please advise as I am unsure.

Many thanks.
Annie
................................
Hi, Annie,

The inverse of the sine is usually written either with an exponent of -1, which is bad because it's not really an exponent here, OR (better) as  'arcsin'.

So y = arcsin(x/5)  does NOT mean what you wrote.  It does mean that:

sin y = x/5

Now use implicit differentiation:

cos y dy/dx  = 1/5

Then you can solve for dy/dx and get:

dy      1
-- = --------
dx   5 cos y

But you don't like having a 'y' in the answer?  No problem -- just use the fact that:

cos^2(y) = 1 - sin^2(y)

cos y = sqrt(1 - sin^2(y))

and since  sin y = x/5,  sin^2(y) = x^2/25

so  cos y = sqrt(1 - x^2/25), and you have:

dy          1
-- = ------------------
dx   5 sqrt(1 - x^2/25)

Now a little algebra:

dy            1
-- = -------------------------
dx   sqrt(25) sqrt(1 - x^2/25)


dy          1
-- = ----------------
dx   sqrt(25 - x^2)