(x^3+yx^3)xy=x^5-y^5
(x^3+yx^3)xy=x^5-y^5
Hmmm. I'd pull out that x^3 first, giving
(1+y)x^4y=x^5-y^5
(y')(x^4y)+(1+y)(4x^3y+x^4y') = 5x^4-5y^4y'
Now expand some to get the y' terms:
x^4yy' + (1+y)(4x^3y) + (1+y)(x^4y') = 5x^4-5y^4y'
y'(x^4y + (1+y)x^4 + 5y^4) = 5x^4 - (1+y)(4x^3y)
Now simplify and divide for
y' =
x^3(5x-4y^2-4y)
-----------------------
2x^4y + x^4 + 5y^4
5x^4-4x^3,y-y^4/x^4+4xy^3+5y^4 answer
http://www.wolframalpha.com/input/?i=derivative+(x%5E3%2Byx%5E3)xy%3Dx%5E5-y%5E5
To solve the equation (x^3+yx^3)xy=x^5-y^5, let's break it down step by step.
Step 1: Distribute the terms on the left side of the equation
(x^3+yx^3)xy = x^5 - y^5
Expanding the first term:
(x^3 + yx^3)xy = (x^3 * xy) + (yx^3 * xy)
Step 2: Simplify the left side
(x^3 * xy) + (yx^3 * xy) = x^4y^2 + x^2y^4
Step 3: The equation becomes
x^4y^2 + x^2y^4 = x^5 - y^5
Step 4: Rearrange the equation to get all terms on one side
x^4y^2 + x^2y^4 - x^5 + y^5 = 0
Step 5: Factor out common terms
x^4y^2 - x^5 + x^2y^4 + y^5 = x^4(y^2 - x) + y^5(x^2 + 1) = 0
Now, we have factored the equation which can help us find any possible solutions for x and y.