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Mathematics 7 Online
OpenStudy (anonymous):

lnsqrt (4 x - 5/ 9 x + 9) Find f'( x )

OpenStudy (anonymous):

quotient rule for this one

OpenStudy (anonymous):

use \[(\frac{f}{g})'=\frac{gf'-fg'}{g^2}\] with \[f(x)=4x-5, f'(x)=4, g(x) = 9x+9, g'(x)=9\]

OpenStudy (anonymous):

oh my i ignored the square root part!

OpenStudy (anonymous):

ya lol i kind of thought soo, but i didnt say anything cuz i didnt know how to do it anyways

OpenStudy (anonymous):

well in that case you still need the derivative of the inside piece

OpenStudy (anonymous):

which comes out to \[\frac{1}{(1+x)^2}\]

OpenStudy (anonymous):

so by the chain rule you will have to take the derivative of the square root function first. \[\frac{d}{dx}[\sqrt{f(x)}=\frac{f'(x)}{2\sqrt{f(x)}}\]

OpenStudy (anonymous):

so you should get \[\frac{\sqrt{9x+9}}{2\sqrt{{4x-5}}}\times \frac{1}{(1+x^2)}\]

OpenStudy (anonymous):

can factor out a 3 in the top

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