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lnsqrt (4 x - 5/ 9 x + 9) Find f'( x )
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quotient rule for this one
use \[(\frac{f}{g})'=\frac{gf'-fg'}{g^2}\] with \[f(x)=4x-5, f'(x)=4, g(x) = 9x+9, g'(x)=9\]
oh my i ignored the square root part!
ya lol i kind of thought soo, but i didnt say anything cuz i didnt know how to do it anyways
well in that case you still need the derivative of the inside piece
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which comes out to \[\frac{1}{(1+x)^2}\]
so by the chain rule you will have to take the derivative of the square root function first. \[\frac{d}{dx}[\sqrt{f(x)}=\frac{f'(x)}{2\sqrt{f(x)}}\]
so you should get \[\frac{\sqrt{9x+9}}{2\sqrt{{4x-5}}}\times \frac{1}{(1+x^2)}\]
can factor out a 3 in the top
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