Asked by susan
dy/dx of the following:
x^2y^2=2/x^2 [ans:-1/x^4y -y/x].
i need the calculation work.
x^2y^2=2/x^2 [ans:-1/x^4y -y/x].
i need the calculation work.
Answers
Answered by
Reiny
I would cross-multiply first
x^4y^2 = 2
now the product rule
x^4(2y)dy/dx + y^2(4x^3) = 0
dy/dx = -4x^3y^2 / ((2x^4)(y))
or directly ...
x^2(2y)dy/dx + 2xy^2 = -4/x^3
dy/dx = (-4/x^3 - 2xy^2)/(2x^2y)
= -2/(x^5y) - y/x
x^4y^2 = 2
now the product rule
x^4(2y)dy/dx + y^2(4x^3) = 0
dy/dx = -4x^3y^2 / ((2x^4)(y))
or directly ...
x^2(2y)dy/dx + 2xy^2 = -4/x^3
dy/dx = (-4/x^3 - 2xy^2)/(2x^2y)
= -2/(x^5y) - y/x
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