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Implicit Differentiation

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Regard y as the independent variable and x as the dependent variable and use implicit differentiation to find dx/dy.

(I attempted this problem multiple times and still can't find what I'm doing wrong)

$$x^5y^2-x^3y+5xy^3=0$$

Mar 7, 2019

#2
+23140
+2

Regard y as the independent variable and x as the dependent variable and use implicit differentiation to find dx/dy.

x^5y^2-x^3y+5xy^3=0

$$\begin{array}{|l|rcll|} \hline & \mathbf{x^5y^2-x^3y+5xy^3} & \mathbf{=}& \mathbf{0} \\\\ \hline y'=\ ? & (5x^{5-1}\cdot y^2+ x^5\cdot 2y^{2-1}\cdot y') \\ & - (3x^{3-1}\cdot y+ x^3\cdot y') \\ & + 5(1\cdot x^{1-1}\cdot y^3 +x\cdot 3y^{3-1}\cdot y' ) &=& 0 \\ \hline & (5x^4y^2+ x^5\cdot 2yy') - (3x^2y+ x^3y') + 5(y^3 +x\cdot 3y^2y' ) &=& 0 \\ & 5x^4y^2+ 2x^5yy'- 3x^2y - x^3y' + 5y^3 +15xy^2y' &=& 0 \\ & 2x^5yy'- x^3y'+15xy^2y'&=& 3x^2y-5y^3-5x^4y^2 \\ & y'x(2x^4y-x^2+15y^2)&=& y(3x^2-5y^2-5x^4y) \\ & \mathbf{y'} &\mathbf{=}& \mathbf{\dfrac{y(3x^2-5y^2-5x^4y)}{x(2x^4y-x^2+15y^2)}} \\ \hline \end{array}$$

Mar 7, 2019
#3
+104417
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Thanks Heureka :)

Melody  Mar 7, 2019
#4
+7712
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$$x^5y^2 - x^3 y + 5xy^3 = 0\\ 5x^4 y^2 \dfrac{dx}{dy} + 2x^5y-3x^2y\dfrac{dx}{dy} -x^3 +5y^3 \dfrac{dx}{dy} + 15xy^2 = 0\\ (5x^4y^2 -3x^2y+5y^3)\cdot\dfrac{dx}{dy} = x^3-2x^5y-15xy^2 \\ \dfrac{dx}{dy} = \dfrac{x(x^2 - 2x^4y-15y^2)}{y(5x^4y-3x^2+5y^2)}$$

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Mar 9, 2019
#5
+23140
+1

Regard y as the independent variable and x as the dependent variable and use implicit differentiation to find dx/dy.

x^5y^2-x^3y+5xy^3=0

heureka  Mar 11, 2019