How do you solve this implicit differentation problem?

Question

18 = e^y + xy^2

Answer 1

18 = e^y + xy^2

0 = e^ydy + y^2dx + 2xydy

dy/dx = -y^2/(e^y + 2xy)

Answer 2

You did no longer extremely say what the question is. if the question could be to precise the by-product is of y with admire to x, then you definately differentiate each little thing interior the equation above: d/dx (x^2y^2) - d/dx(2xy^2) = 0 2x y^2 + 2x^2 y dy/dx - 2 y^2 -4 xy dy/dx = 0 we are able to divide all by utilising 2. then we get: xy^2 - y^2 = (2xy - x^2 y) dy/dx dy/dx = (xy^2 - y^2) / (2xy - x^2 y) (are you able to thrill be sure?)

Answer 3

18 = e^y + xy^2

Differentiating both sides respect to x:

D(18) = D(e^y + xy^2),

0 = D(e^y) + D(xy^2),

0 = D(e^y) + D(x) y^2 + xD(y^2) ,

0 = (e^y)y’ + y^2 + 2yy’x,

Solving for y’,

y’ = -y^2 / (e^y + 2xy )

Answer 4

Differentiate:
18 => 0
e^y => e^y dy
now with xy^2, you need to use the product rule
differentiate with respect to x, then with respect to y.
xy^2 => y^2 dx + x *2y dy

0 = e^y dy + y^2 dx + x*2y dy group by dy, and move the dx over
-y^2 dx = (e^y + 2xy)dy divide out
dx/dy = -(e^y + 2xy)/y^2
or
dy/dx = -y^2/(e^y + 2xy)

Answer 5

first... differ each variable
(D/DX) = ...... y^2

(D/DY) = y e^y + 2xy

(dy/dx) = (- y^2) / y e^y + 2xy