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

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Answer 1 · 2016-10-15T16:49:03

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

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

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 $8x^3$ :

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

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

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

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