# Find dy/dx if x^5 + 4xy^3 – y^5 =2

Asked on by michdon19

neela | High School Teacher | (Level 3) Valedictorian

Posted on

To find dy/dx if x^5 + 4xy^3 – y^5 = 2.

The relation between x and y in this equation is implicit. In such cases we straight away differentiate term by terrm and  try to get dy/dx by solving for dy/dx  from the equation.

Differentiating both sides of the given equation with respect to x, we get:

(x^5 + 4xy^3 – y^5)' =(2)'

(x^5)' + {4xy^3}' – {y^5}' = 0

{x

5x^4 +{4(x)'y^3 +4x(y^3)'}-{5y^4*dy/dx} = 0.

5x^4+4y^3 +4x*3y^2dy/dx-5y^4dy/dx = 0. We try to solve for dy/dx from this equation:

5x^x+4y^3 +{12x^3y^2 - 5y^4) dy/dx = 0

(12x^3y^2-5y^4)dy/dx.

Divide both sides by the coefficient of dy/dx, that is, 12x^2-5y^4.

dy/dx = -(5x^4+4y^3)/(12x^3y^2-5y^4). Or

dy/dx = (4y^3+5x^4)/(5y^4-12x^3y^4)

william1941 | College Teacher | (Level 3) Valedictorian

Posted on

Here we are given the equation x^5 + 4xy^3 – y^5 =2 to differentiate.

We don’t need to express y in terms of x to do this, we can differentiate straight away.

d/dx(x^5 + 4xy^3 – y^5) = d/dx (2)

=> d/dx (x^5) + d/dx (4xy^3) – d/dx (y^5) = d/dx (2)

=> 5x^4 + 4x*y^2 dy/dx + 4y^3 – 5y^4 dy/dx = 0

=> 5x^4 + dy/dx [4xy^2 – 5y^4] + 4y^3=0

=> dy/dx [4xy^2 – 5y^4] = - [4y^3 + 5x^4]

=> dy/dx = - [4y^3 + 5x^4]/ [5y^4 - 4xy^2]

Therefore the result is dy/dx = [4y^3 + 5x^4]/ [5y^4 - 4xy^2]

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