What is the derivative of #sqrt(x)/(x^3+1)# using the quotient rule?

1 Answer
Apr 11, 2015

We'll use the quotient rule, and, because the square root is used so often (for example: diagonal distance = hypotenuse of a right triangle), it's nice to memorize:

#d/(dx)sqrtx = 1/(2sqrtx)# (Go through it witl exponents to see that it's true.)

So:

#d/(dx)(sqrt(x)/(x^3+1)) = (1/(2sqrtx)(x^3+1) - (sqrtx) (3x^2))/((x^3+1)^2)#

Mulhtiply by #(2sqrtx)/(2sqrtx)# to get:

#d/(dx)(sqrt(x)/(x^3+1)) = ((x^3+1) - 2x (3x^2))/((2sqrtx)(x^3+1)^2) = (-5x^2+1)/((2sqrtx)(x^3+1)^2) #