What is the derivative of #y=(x^2lnx)^4#?

1 Answer
Aug 14, 2015

#y^' = 4x^7 * ln^3(x) * (2ln(x) + 1)#

Explanation:

Notice that you can simplify your function to get

#y = (x^2)^4 * (lnx)^4#

#y = x^8 * ln^4(x)#

To differentiate this function you can use the product rule and the chain rule for #u^4#, with #u = ln(x)#.

#d/dx(y) = [d/dx(x^8)] * ln^4(x) + x^8 * d/dx(ln^4(x))#

#y^' = 8x^7 * ln^4(x) + x^8 * ([d/(du)(u^4)] * d/dx(u))#

#y^' = 8x^7 * ln^4(x) + x^color(red)(cancel(color(black)(8))) * 4 ln^3(x) * 1/color(red)(cancel(color(black)(x)))#

#y^' = 8x^7 * ln^4(x) + 4x^7 * ln^3(x)#

This can be simplified to give

#y^' = color(green)(4x^7 * ln^3(x) * (2ln(x) + 1))#