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

1 Answer
Oct 24, 2017

Answer:

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

Explanation:

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)#