If f(x)=6arcsin(x^3), find f'(0.5)=?

Answer 1

An identity you should remember is d/dx(arcsin(x)) = 1/sqrt(1 - x^2). Other derivative inverse trig identities include d/dx(arcsec(x)) = 1/(abs(x)*sqrt(x^2-1)) and d/dx(arctan(x)) = 1/(x^2 + 1). I think the last one has come up the most for me in other problems. The derivatives of the inverses of the cofunctions (cos, csc, cot) are the same as for the regular functions but negated.

That being said, to compute f '(x) we first use the chain rule. Carry the constant out front along for the ride:

f '(x) = 6* d/du(arcsin(u))* 3x^2 = 18x^2 * 1/sqrt(1 - u^2) = 18x^2/sqrt(1 - x^6).

where u = x^3. Plug in 0.5 to above to find f '(0.5).

Answer 2

f'(x) = 6* 3x^2/sqrt ( 1 - (x^3))^2 )

= 18*x^2 / sqrt ( 1 - x^6)

f'(0.5) = 4.5 / sqrt ( 1 - 0.5^6)

= 4.5 / sqrt( 0.984375)

= 4.5355736761107267....