The derivative of arccos(x) is so the derivative of g is . Now take that "2"insidethe squareroot to get .

Multiply both numerator and denominator of your answer by .2. f(x) = arctan (x/a)

I got:

[1/a] / [1 + x^2/a^2]

The answer is:

a / [a^2+x^2]

Use the quotient rule: .3. g(x) = arcsin (3x) / x

NO IDEA

The answer is:

[3x - sqrt(1-9x^2)*arcsin(3x)] / [x^2*sqrt(1-9x^2)]

The best way to do this is to note that is y= arccos(t), which is the same as t= cos(y), then we can imagine a right triangle with angle y, hypotenuse of length 1 and "opposite side" t. By the Pythagorean theorem, the "near side" has length and . So . Use the chain rule to differentiate that:4. h(t) = sin[arccos(t)]

The answer is:

-t/sqrt(1-t^2)

Use the product rule for x arccos(x): (x)' arccos(x)+ x(arccos(x))'= arccos(x)- while the derivative of we just did above- its derivative is . You see that cancels the previous term, leaving only arccos(x).5. y = x*arccos(x) - sqrt(1-x^2)

The answer is:

arccos(x)

The first thing I would do is write ln((x+1)/(x-1)) as ln(x+1)- ln(x-1) so its derivative is . Of course the derivative of arctan(x) is .6. y = (1/2) * [ (1/2)*ln{x+1/X-1} + arctan(x) ]

The answer is:

1/[1-x^4]

y'= . (I've dropped the 1/2 multiplying the whole thing so I won't have to carry it along.) Add those fractions by getting the common denominator

.

Putting back the 1/2 I dropped gives .