How do you differentiate $y=ln(9x+4)^2$ ?

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Answer 1 · 2017-10-24T16:30:32

$dy/dx=(18ln(9x+4))/(9x+4)$

The chain rule states that $dy/dx=dy/(du)(du)/dx$ for some function $u$ of $x$ .

In this problem, let $u=ln(9x+4)$ , so $y=u^2$ . Therefore, $dy/dx=d/(du)[u^2]times(du)/dx$
$dy/dx=2utimes(du)/dx$

And substitute back in for $u$ .
$dy/dx=2ln(9x+4)times(du)/dx$

So now we need to find $(du)/dx$ , we can use the chain rule again. Now say $(du)/dx=(du)/(dv)(dv)/dx$

Let $v=9x+4$ so $u=lnv$
$(du)/dx=d/(dv)[lnv]times(dv)/dx$
$(du)/dx=1/vtimes(dv)/dx$

And substitute back in for $v$ .
$(du)/dx=1/(9x+4)times(dv)/dx$

And finally we need to find $(dv)/dx$ .
$(dv)/dx=d/dx[9x+4]$
$(dv)/dx=9$

And finally plug everything back in.
$(du)/dx=1/(9x+4)times9=9/(9x+4)$
$dy/dx=2ln(9x+4)times9/(9x+4)=(18ln(9x+4))/(9x+4)$

How do you differentiate y=ln(9x+4)^2y=ln(9x+4)^2?