How do you use the chain rule to differentiate #2^(ln4x)#?

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Answer 1 · 2016-10-11T01:12:18

This type of problem is done using logarithmic differentiation.

Answer: #(dy)/dx= ln(2)(2^(ln(4x)))/x#

This type of problem is done using logarithmic differentiation:

Let #y = 2^(ln(4x))#

Take the natural log of both sides:

#ln(y) = ln(2^(ln(4x)))#

Use the identity #ln(a^b) = ln(a)b# on the right side:

#ln(y) = ln(2)(ln(4x))#

Differentiate the left side:

#(d[ln(y)])/dx = (1/y)((dy)/dx)#

Differentiate the right side:

#ln(2)# is just a constant; it is #ln(4x)# that requires the chain rule:

Let #u(x) = 4x#, then, #(dln(u))/(du) = 1/u# and #(du)/dx = 4#

The chain rule is:

#(d[ln(u(x))])/dx = ((d[ln(u)])/(du))((du)/dx)#

#(d[ln(u(x))])/dx = (1/u)(4)#

Don't forget the constant, ln(2):

#ln(2)(d[ln(4x)])/dx = ln(2)(1/u)(4)#

Reverse substitution for u:

#ln(2)(d[ln(4x)])/dx = (4ln(2))/(4x)#

#ln(2)(d[ln(4x)])/dx = ln(2)/x#

Put the left and right sides back together:

#(1/y)((dy)/dx)= ln(2)/x#

Multiply both sides by y:

#(dy)/dx= ln(2)y/x#

Substitute #2^(ln(4x))# for #y#:

#(dy)/dx= ln(2)(2^(ln(4x)))/x#