What is the derivative of #(4x^8-sqrt(x))/(8x^4)#?

1 Answer
Aug 1, 2015

#y^' = 1/16 * (32x^(15/2) + 7) * x^(-9/2)#

Explanation:

Start by rewriting your function like this

#y = (4x^color(red)(cancel(color(black)(8))))/(8color(red)(cancel(color(black)(x^4)))) - x^(1/2)/(8x^4)#

#y = 1/2x^4 - 1/8 * x^(-7/2)#

Now you can use the power rule to differentiate #y#

#y = 1/2[d/dx(x^4)] - 1/8 d/dx(x^(-7/2))#

#y = 1/color(red)(cancel(color(black)(2))) * color(red)(cancel(color(black)(4))) x^3 - 1/8 * (-7/2) * x^(-9/2)#

#y = 2x^3 + 7/16x^(-9/2)#

This can be rewritten as

#y^' = (x^(-9/2) * (16 * 2x^(15/2) + 7))/16 = color(green)(1/16 * (32x^(15/2) + 7) * x^(-9/2))#