How do you find the derivative of the function?

Question

1)f(x)=arcsinx+arccosx
2) g(x)=(arcsin3x)/x

Answer 1

f(x) = arcsin(x) + arccos(x)

The derivative of arcsin(x) is 1/sqrt(1 - x^2) and the deriative of arccos(x) is -1/sqrt(1 - x^2).

Therefore,

f'(x) = 1/sqrt(1 - x^2) + -1/sqrt(1 - x^2)
They are negations of each other, and thus
f'(x) = 0

{This implies arcsin(x) + arccos(x) is a constant.}

2) g(x) = arcsin(3x) / x

Using the product rule,

g'(x) = { [1/sqrt(1 - x^2)](3)[x] + arcsin(3x) } / x^2

To convert this from a complex fraction to a simple fraction, multiply top and bottom by sqrt(1 - x^2).

g'(x) = [ 3x + sqrt(1 - x^2) arcsin(3x) ] / [x^2 sqrt(1 - x^2) ]

From here we can rationalize the denominator and all that good stuff, but simplification techniques are second to knowing how to evaluate the derivative.

Answer 2

1. f(x) = arcsin x + arccos x

f'(x) = 1/ sqrt(1-x^2) + - 1/sqrt(1- x^2)

f'(x) = 0

2. g(x) = (arcsin 3x)/x

g'(x) = (x(1/sqrt(1-(3x)^2))(3) - arcsin 3x)/ x^2

g'(x) = (3x/sqrt(1-9x^2) - arcsin 3x)/ x^2

sorry, it looks a little wierd... it really looks better on paper....