`x^3 + y^3 = 6xy - 1, (2,3)` Find `dy/dx` by implicit differentiation and evaluate the derivative at the given point.

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Textbook Question

Chapter 2, 2.5 - Problem 26 - Calculus of a Single Variable (10th Edition, Ron Larson).
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mathace | (Level 3) Assistant Educator

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Given: `x^3+y^3=6xy-1,(2,3)`

`3x^2+3y^2dy/dx=6xdy/dx+y(6)`

`3y^2dy/dx-6xdy/dx=6y-3x^2`

`dy/dx(3y^2-6x)=6y-3x^2`

`dy/dx=(6y-3x^2)/(3y^2-6x)`

`dy/dx=[3(2y-x^2)]/[3(y^2-2x)]`

`dy/dx=(2y-x^2)/(y^2-2x)`

Evaluate the derivative by substituting the (2, 3) into the derivative.

`dy/dx=[2(3)-(2)^2]/[(3)^2-2(2)]=2/5`

gsarora17 | (Level 2) Associate Educator

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`x^3+y^3=6xy-1`

Differentiating both sides with respect to x,

`3x^2+3y^2dy/dx=6(xdy/dx+y)`

`3x^2+3y^2dy/dx=6xdy/dx+6y`

`(3y^2-6x)dy/dx=6y-3x^2`

`dy/dx=(6y-3x^2)/(3y^2-6x)`

`dy/dx=(2y-x^2)/(y^2-2x)`

Now derivative at the point (2,3) can be obtained by plugging in the values of x and y in the dy/dx.

Derivative at (2,3)=`(2*3-2^2)/(3^2-2*2) = 2/5`

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gsarora17 | (Level 2) Associate Educator

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Not able to understand how the final value of derivative =2/5 did not appeared.

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