How do you differentiate $y= ln (ln6x)$ ?

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Answer 1 · 2016-04-02T20:28:10

$dy/dx = 1/(xln(6x))$

You will have to use the chain rule twice, as there is one embedded function inside another: $6x$ inside $ln$ inside another $ln$ .

The chain rule states that

$f(x) = g(h(x))$
$f'(x) = h'(x)g'(h)$

Let's solve for the internal $ln(6x)$ first.

$h'(x) = d/dx 6x = 6$
$g'(h) = 1/(h(x)) = 1/(6x)$
$d/dx ln(6x) = 6/(6x) = 1/x$

That is just the inside function. The derivative of the whole thing will need to use the chain rule again.

$f'(x) = h'(x)g'(h)$
$h'(x) = 1/x$
$g'(h(x)) = 1/(h(x)) = 1/ln(6x)$

$h'(x)g'(h) = 1/x * 1/ln(6x) = 1/(xln(6x))$

How do you differentiate y= ln (ln6x)y= ln (ln6x)?