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Help with quotient/chain rule, please, I'm stuck!!?

+2 votes


What is the derivative of 
((sin^-1)(x^1/2))
------------
(1-x)^1/2

Any help would be amazing, thank you!

asked Jan 29, 2013 in CALCULUS by andrew Scholar

1 Answer

+4 votes

= (sin-1√x) / [√(1 - x)]

Derivative formulas:

Recall: quotient / chain rule f(x)·g(x) = [ f'(x)·g(x) - f(x)·g'(x) ] / g2(x)

The Derivative of the Inverse Sine. (d/dx)(sin-1x) = 1/√(1-x2)

The Power Rule (Variable raised to a constant). (d/dx)(An) = n An-1 and (d/dx)(√x) = 1/(2√x)

= [(sin-1√x)'·√(1 - x) - (sin-1√x)·√(1 - x)' ] / [√(1 - x)]2

= {[1/√(1-√x2)][1/(2√x)][√(1 - x)] - (sin-1√x) [1/(2√1-x)](-1)} / (1 - x)         (inner derivative)

= {[1/√(1- x)][1/(2√x)][√(1 - x)] + (sin-1√x) [1/(2√1-x)]} / (1 - x)                         (Simplify)

= {[1/(2√x)] + (sin-1√x) [1/(2√1-x)]} / (1 - x)

Take out denominator terms.

= [1/2(1-x)][1/(√x) + (sin-1√x) / (√1-x)]

There fore the derivative of (sin-1√x) / [√(1 - x)] = [1/2(1-x)][1/(√x) + (sin-1√x) / (√1-x)].

 
answered Jan 31, 2013 by richardson Scholar

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