How do you differentiate #y=ln ln(2x^4)#?

Question

Charles Arocha · Accepted Answer

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

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

Through the chain rule, we see that:

#d/dxln(color(red)f(x))=1/color(red)f(x)*d/dxcolor(red)f(x)#

So, here, we see that:

#dy/dx=d/dxln(color(red)ln(2x^4))=1/color(red)ln(2x^4)*d/dxcolor(red)ln(2x^4)#

Note that we will use the same rule again to find #d/dxln(2x^4)#:

#dy/dx=1/ln(2x^4) * [d/dxln(color(red)(2x^4))]=1/ln(2x^4) * [1/color(red)(2x^4)*d/dxcolor(red)(2x^4)]#

Differentiating #2x^4# through the power rule gives #8x^3#:

#dy/dx=1/ln(2x^4) * 1/(2x^4) * 8x^3#

#dy/dx=1/ln(2x^4) * 1/x * 4#

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

Luke Alvarez · Answer

undefined To differentiate $ y = \ln(\ln(2x^4)) $, we use the chain rule:

1. First, differentiate the outer function, ln(u), where u = ln(2x^4), which gives 1/u.
2. Then, multiply by the derivative of the inner function, u = ln(2x^4), which involves applying the chain rule again.

Derivative of the inner function u:
$$ u = \ln(2x^4) $$
$$ u' = \frac{1}{u} \cdot \left(\frac{1}{2x^4}\right) $$

Now, putting it all together:
$$ y' = \frac{1}{\ln(2x^4)} \cdot \frac{1}{2x^4} $$

So, the derivative of $ y = \ln(\ln(2x^4)) $ is:
$$ y' = \frac{1}{\ln(2x^4)} \cdot \frac{1}{2x^4} $$