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Calculus/Inv Trig Differentiation

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Expert: - 12/6/2008


Question
Trying to differentiate Arcsin (sqrt (1-x )). Can you help me? Do I need all the square roots?  I got this problem from this site (see no. 12).

http://www.maths.abdn.ac.uk/~igc/tch/ma1002/diff/node47.html  

Answer
Here is a note: dy/dx is really a fraction.
If y=x², dy/dx = 2x dx, so dy = 2x dx.
If y=4x^3, dy/dx = 12x², so dy = 12x² dx.
Noe that we have that down, lets proceed.

Let's start with y=Arscis(√(1-x)).
What can be said is that sin(y) = √(1-x).

We can differentiate both sides, giving
cos(y)dy = -1/[2√(1-x)] dx.

Divide both sides by dx and cos(y) and note that 1/cos(y) = sec(y).
[1] dy/dx = -sec(y)/[2√1-x)] dx.

What needs to be done now is to find sec(y) in terms of x.

Take a right triangle with opposite side b = √(1-x)
and hypoteneuse c=1.  The near side can be found to be a=√(c²-b²).

The value of sec(y) can be replaced with c/a.  The values for c and a were given in the preceeding paragraph.  Take equation [1] and replace sec(y) with c/a.

Now you have a formula for dy/dx in terms of x alone.

Note: The equation      cos(y)dy = -1/[2√(1-x)] dx

is what is referred to as a simple differential equation.

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