How do you find the derivative of #f(x)= 1/(9x+6)^2#?

1 Answer
Feb 22, 2016

Answer:

#f'(x) = -2 / (3*(3x + 2)^{3})#

Explanation:

Since this is under the chain rule, then use it.

Substitute #u = 9x + 6#.

#frac{"d"u}{"d"x} = 9#

So plug it in.

#f'(x) = frac{"d"}{"d"x}(1/(9x + 6)^2)#

#= frac{"d"}{"d"x}(1/u^2)#

#= frac{"d"}{"d"u}(u^{-2})*frac{"d"u}{"d"x}#

#= (-2) * u^{-3} * (9)#

#= -18 / (9x + 6)^{3}#

#= -18 / (3^3*(3x + 2)^{3})#

#= -2 / (3*(3x + 2)^{3})#