How do you find the derivative of #(x^2 + 6)^(−1/7) −( 2/7)x^2(x^2 + 6)^(−8/7)#?

1 Answer
Aug 4, 2015

Answer:

#f^' = -(2x)/49 * (5x^2 + 126)/(x^2+6)^(15/7)#

Explanation:

You're going to have to use the power, chain, and product rules to differentiate this function.

To make the calculations easier to follow, I'll write the function as the sum of two other functions and calculate their derivatives separately.

#f(x) = g(x) + h(x)#, where

#g(x) = (x^2 + 6)^(-1/7)#
#h(x) = -2/7x^2(x^2 + 6)^(-8/7)#

So, start with the derivative of #g(x)#, for which you can use the chain and power rules.

#d/dx(g(x)) = d/dx(x^2 + 6)^(-1/7)#

#g^' = -1/7 * (x^2 + 6)^(-1/7 - 1) * d/dx(x^2 + 6)#

#g^' = -1/7 * (x^2 + 6)^(-8/7) * 2x#

Next, determine the derivative of #h(x)#, for which you can use the power, chain, and product rules.

#d/dx(h(x)) = -2/7 [d/dx(x^2) * (x^2 + 6)^(-8/7) + x^2 * d/dx(x^2 + 6)^(-8/7)]#

#h^' = -2/7[ 2x * (x^2 + 6)^(-8/7) + x^2 * (-8/7 * (x^2 + 6)^(-15/7) * 2x)]#

#h^' = -2/7[2x * (x^2 + 6)^(-8/7) - (16x^3)/7(x^2 + 6)^(-15/7)]#

For simplicity, I'll use the notation #(x^2 + 6) = a# until I can get the target derivative to a more compact form.

#f^' = g^' + h^'#

#f^' = -(2x)/7 * a^(-8/7) - (4x)/7 * a^(-8/7) + (32x^3)/49 * a^(-15/7)#

#f^' = -(6x)/7a^(-8/7) + (32x^3)/49 * a^(-15/7)#

You can simplify this further to get

#f^' = -(2x)/7a^(-8/7) * (3 - (16x^2)/7 * a^(-1))#

#f^' = -(2x)/7a^(-8/7) * (21a - 16x^2)/(7a)#

FInally, replace #a# with #x^2 + 6# to get

#f^' = (-2x)/7 * (21(x^2 + 6) - 16x^2)/(7 * (x^2 + 6)^(15/7))#

#f^' = -(2x)/7 * (5x^2 + 126)/(7(x^2+6)^(15/7))#

#f^' = color(green)(-(2x)/49 * (5x^2 + 126)/(x^2+6)^(15/7))#