How do you use the chain rule to differentiate #ln(4x)^10#?

1 Answer
Jim H
Dec 14, 2016

Please see below.

Explanation:

My preference would not be the chain rule, but here's how to differentiate using the chain rule.

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

#d/dx(ln(4x)^10) = d/dx(ln(underbrace((4x)^10)_u))#

# = 1/(underbrace((4x)^10)_u) * d/dx(underbrace((4x)^10)_u)#

# = 1/(4x)^10 * 10(4x)^9 * d/dx(4x)#

# = 1/(4x)^10 * 10(4x)^9 * 4#

# = (40(4x)^9)/(4x)^10#

# = 4/(4x) = 1/x#.

Using properties of logarithms to rewrite first

#ln(4x)^10 = 10 ln(4x) = 10(ln4+lnx) = 10ln4+10lnx#.

Note that #10ln4# is a constant, so its derivative is #0#.

#d/dx(ln(4x)^10) = d/dx(10ln4+10lnx)#

# = 0+10*1/x#

# = 10/x#.

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