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Find dy/dx if y= 5x^3+2/x^2+4 +6
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dy/dx =15x^2- 4/X^3 (not sure if im right though xD)
IS that 6 outside the fraction?
yes
\[y =\frac{5x^3 + 2 }{x^2+4}+6\]
it this the equation
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yes
no
5x^3 is not in the fraction
\[y = 5x^3 + \frac{2}{x^2 +4} +6\]
yes
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lols , is that the question damn i got it wrong then argh
mimi try again
lols , ok
\[\frac{dy}{dx} = 10x^2 - \frac{2(2x)}{(x^2+4)^2} \]
15x^2????? explain pls
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isn't derivative of 5x^3 , 15x^2
yeah 15x^2 sorry
explain the fraction part
why multiply 2x
\[\frac{d}{dx}\frac{f(x)}{g(x)} = \frac{f'(x) g(x)- g'(x)f(x)}{(g(x))^2}\]
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use this you will get the answer
okay thanks
i still have another question if you dont mind please give full solution with explanation thanks a bundle
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