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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Answer 1 · 2016-08-05T14:49:39

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

Explanation:

When we implicitly differentiate a function $f (x, y) = 0$ $f(x,y)=0$ , whenever we differentiate w.r.t. we use chain rule and when we differentiate w.r.t. $y$ $y$ and then multiply it by $\frac{d y}{d x}$ $(dy)/(dx)$ .

As such as we have $3 y + \frac{y^{4}}{x^{2}} = 2$ $3y+y^4/x^2=2$

$3 \times 1 \times \frac{d y}{d x} + \frac{4 y^{3} \times \frac{d y}{d x} - y^{4} \times 2 x}{x^{4}} = 0$ $3xx1xx(dy)/(dx)+(4y^3xx(dy)/(dx)-y^4xx2x)/x^4=0$ or

$3 \frac{d y}{d x} + \frac{4 y^{3} \frac{d y}{d x} - 2 x y^{4}}{x^{4}} = 0$ $3(dy)/(dx)+(4y^3(dy)/(dx)-2xy^4)/x^4=0$ or

$3 \frac{d y}{d x} + 4 \frac{y^{3}}{x^{4}} \frac{d y}{d x} - 2 \frac{y^{4}}{x^{3}} = 0$ $3(dy)/(dx)+4y^3/x^4(dy)/(dx)-2y^4/x^3=0$ or

$\frac{d y}{d x} [3 + \frac{4 y^{3}}{x^{4}}] = \frac{2 y^{4}}{x^{3}}$ $(dy)/(dx)[3+(4y^3)/x^4]=(2y^4)/x^3$ or

$\frac{d y}{d x} = \frac{\frac{2 y^{4}}{x^{3}}}{3 + \frac{4 y^{3}}{x^{4}}}$ $(dy)/(dx)=((2y^4)/x^3)/(3+(4y^3)/x^4)$ or

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

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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!

1 Answer

Explanation:

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How do you implicitly differentiate 3y + y^4/x^2 = 23y+y4x2=23y + y^4/x^2 = 2?

Thanks for your help!

1 Answer

Explanation:

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