What is (\partial)/(\partial y)(x^2-xy+4y^2) ?

The answer to (\partial)/(\partial y)(x^2-xy+4y^2) is 8y-x

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(\partial)/(\partial y)(x^2-xy+4y^2)

\frac{\partial}{\partial y} (x^{2} - x y + 4 y^{2})

8 y - x

Solution steps

\frac{\partial}{\partial y} (x^{2} - x y + 4 y^{2})

Treat

x

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{\partial}{\partial y} (x^{2}) - \frac{\partial}{\partial y} (x y) + \frac{\partial}{\partial y} (4 y^{2})

\frac{\partial}{\partial y} (x^{2}) = 0

\frac{\partial}{\partial y} (x y) = x

\frac{\partial}{\partial y} (4 y^{2}) = 8 y

= 0 - x + 8 y

Simplify

= 8 y - x

x \to 2 lim (2 x - 7)

\frac{d}{d x} (\frac{sin ( 2 x )}{cos ( x )})

\frac{d}{d x} (4^{c o s (x)})

\int \frac{x ^{3}}{( x ^{2} + 1 ) ^{2}} d x

x \to - 5 lim (\frac{x - 5}{x + 5})

What is (\partial)/(\partial y)(x^2-xy+4y^2) ?
The answer to (\partial)/(\partial y)(x^2-xy+4y^2) is 8y-x