How do you implicitly differentiate $3 y + \frac{y^{4}}{x^{2}} = 2$ $3y + y^4/x^2 = 2$ ?

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How do you implicitly differentiate $3 y + \frac{y^{4}}{x^{2}} = 2$ $3y + y^4/x^2 = 2$ ?

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2 Answers

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Answer 1 · 2016-07-07T17:08:18

I found: $\frac{d y}{d x} = \frac{2 y^{4}}{3 x^{3} + 4 x y^{3}}$ $(dy)/(dx)=(2y^4)/(3x^3+4xy^3)$

Explanation:

ou need to remember to derive also $y$ $y$ as a function of $x$ $x$ , so, for example:
$y^{2}$ $y^2$ derived will give you: $2 y \frac{d y}{d x}$ $2y(dy)/(dx)$
in your case you get:
$3 \frac{d y}{d x} + \frac{4 y^{3} x^{2} \frac{d y}{d x} - 2 x y^{4}}{x^{4}} = 0$ $3(dy)/(dx)+(4y^3x^2(dy)/(dx)-2xy^4)/x^4=0$
$3 \frac{d y}{d x} + 4 \frac{y^{3}}{x^{2}} \frac{d y}{d x} - 2 \frac{y^{4}}{x^{3}} = 0$ $3(dy)/(dx)+4y^3/x^2(dy)/(dx)-2y^4/x^3=0$
collect $\frac{d y}{d x}$ $(dy)/(dx)$ and rearrange:
$\frac{d y}{d x} = \frac{2 \frac{y^{4}}{x^{3}}}{3 + 4 \frac{y^{3}}{x^{2}}} = \frac{2 y^{4}}{3 x^{3} + 4 x y^{3}}$ $(dy)/(dx)=(2y^4/x^3)/(3+4y^3/x^2)=(2y^4)/(3x^3+4xy^3)$

Answer 2 · 2016-07-07T17:13:26

yes

Explanation:

me too when it was asked elsewhere

https://socratic.org/questions/how-do-you-implicitly-differentiate-3y-y-4-x-2-2-2#285710

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How do you implicitly differentiate $3 y + \frac{y^{4}}{x^{2}} = 2$ $3y + y^4/x^2 = 2$ ?

Thanks for your help!

2 Answers

Explanation:

Explanation:

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