Asked by warnar
(x^3+yx^3)xy=x^5-y^5
Answers
Answered by
Steve
(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
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
Answered by
warnar
5x^4-4x^3,y-y^4/x^4+4xy^3+5y^4 answer
Answered by
Steve
http://www.wolframalpha.com/input/?i=derivative+(x%5E3%2Byx%5E3)xy%3Dx%5E5-y%5E5
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