Question

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

Answer 1

This type of problem is done using logarithmic differentiation.

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

Explanation:

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`