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Mathematics 23 Online
OpenStudy (anonymous):

Find dy/dx if y= 5x^3+2/x^2+4 +6

OpenStudy (mimi_x3):

dy/dx =15x^2- 4/X^3 (not sure if im right though xD)

OpenStudy (anonymous):

IS that 6 outside the fraction?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

\[y =\frac{5x^3 + 2 }{x^2+4}+6\]

OpenStudy (anonymous):

it this the equation

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

no

OpenStudy (anonymous):

5x^3 is not in the fraction

OpenStudy (anonymous):

\[y = 5x^3 + \frac{2}{x^2 +4} +6\]

OpenStudy (anonymous):

yes

OpenStudy (mimi_x3):

lols , is that the question damn i got it wrong then argh

OpenStudy (anonymous):

mimi try again

OpenStudy (mimi_x3):

lols , ok

OpenStudy (anonymous):

\[\frac{dy}{dx} = 10x^2 - \frac{2(2x)}{(x^2+4)^2} \]

OpenStudy (anonymous):

15x^2????? explain pls

OpenStudy (mimi_x3):

isn't derivative of 5x^3 , 15x^2

OpenStudy (anonymous):

yeah 15x^2 sorry

OpenStudy (anonymous):

explain the fraction part

OpenStudy (anonymous):

why multiply 2x

OpenStudy (anonymous):

\[\frac{d}{dx}\frac{f(x)}{g(x)} = \frac{f'(x) g(x)- g'(x)f(x)}{(g(x))^2}\]

OpenStudy (anonymous):

use this you will get the answer

OpenStudy (anonymous):

okay thanks

OpenStudy (anonymous):

i still have another question if you dont mind please give full solution with explanation thanks a bundle

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