Question

Calculate dy/dx, (x^3)(y^7) + 16xy^6=0

My ans is (y(-16-3x^2y))/ x(96+7x^2y) but at wolfram is y^6(3x^2y+16). Btw i use product rule to dy/dx.

Need someone to recheck my ans.

Answer 1

3x^2y^7 + 7x^3y^6y' + 16y^6 + 96xy^5 y' = 0
y'(7x^3y^6+96xy^5) = -(3x^2y^7 + 16y^6)

y' =
-y^6 (3x^2 y + 16)
---------------------
xy^5 (7x^2 y + 96)

y' =
-y(3x^2y+16)
-----------------
x(7x^2 y + 96)

you are correct, and Wolfram agrees for me. If the answer you got was different, you must have mistyped it into Wolfram

Answer 2

Thanks guy