15. The equation below defines z as a differentiable function of x and y. Find the value of dz/dy at the point (1, 1, 1).

x² - 5y² + xyz² = y - 4​

15 The equation below defines z as a differentiable function of x and y Find the value of dzdy at the point 1 1 1 x 5y xyz y 4 class=

Respuesta :

Isolate the term with [tex]z^2[/tex].

[tex]x^2 - 5y^2 + xyz^2 = y - 4 \implies xyz^2 = -x^2 + 5y^2 + y - 4[/tex]

Differentiate both sides with respect to [tex]y[/tex].

[tex]\dfrac{\partial(xyz^2)}{\partial y} = \dfrac{\partial(-x^2 + 5y^2 + y - 4)}{\partial y}[/tex]

By the product and chain rules,

[tex]xz^2 + 2xyz \dfrac{\partial z}{\partial y} = 10y + 1[/tex]

Solve for the partial derivative, then evaluate at [tex](x,y,z) = (1,1,1)[/tex].

[tex]\dfrac{\partial z}{\partial y} = \dfrac{10y + 1 - xz^2}{2xyz}[/tex]

[tex]\dfrac{\partial z}{\partial y} \bigg|_{x=1,y=1,z=1} = \dfrac{10 + 1 - 1}{2} = \boxed{5}[/tex]