How do you differentiate #f(x)=ln((2x+5)^5)^2 # using the chain rule?

1 Answer
Nov 14, 2016

Answer:

Rewrite it first.

Explanation:

#f(x)=ln((2x+5)^5)^2 #

The convention for reading this is that we find #2x+5#, then rais that number to the #5^"th"# power, then square that result and finally take the natural logarithm of that.

By propeties of exponents, #(u^5)^2 = u^10#.

By properties of logarithms, #ln(u^10) = 10 lnu#.

Applying the chain rule, we have

#d/dx(lnu) = 1/u (du)/dx#.

#f(x)=ln((2x+5)^5)^2 = 10ln(2x+5)#

#f'(x) = 10 * 1/(2x+5) * d/dx(2x+5)#

# = 20/(2x+5)#