How to differentiate $y= (ln(4x^2))/(3^(6x)$ ?

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Answer 1 · 2018-05-29T14:02:05

$(2-6x ln 3 ln(4x^2))/(3^(6x)x)$

You can use the quotient rule

$d/dx (u/v) = (v (du)/dx - u(dv)/dx)/v^2$

Here

$u = ln(4x^2) = ln 4+2ln x implies$
$(du)/dx = 2/x$

and

$v = 3^(6x) = (e^{ln 3})^{6x} = e^{6 (ln 3 )x}implies$

$(dv)/dx = 6 ln 3 times e^{6 (ln 3 )x} = 6 ln 3 times 3^(6x)$

Hence

$d/dx(ln(4x^2)/3^(6x)) =(3^(6x)times 2/x-ln(4x^2)times 6 ln 3 times 3^(6x))/(3^(6x))^2$
$qquad = (2-6x ln 3 ln(4x^2))/(3^(6x)x)$

How to differentiate y= (ln(4x^2))/(3^(6x)y= (ln(4x^2))/(3^(6x) ?