What is (\partial)/(\partial y)(20(y-x^2)) ?

The answer to (\partial)/(\partial y)(20(y-x^2)) is 20

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(\partial)/(\partial y)(20(y-x^2))

\frac{\partial}{\partial y} (20 (y - x^{2}))

20

Solution steps

\frac{\partial}{\partial y} (20 (y - x^{2}))

Treat

x

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 20 \frac{\partial}{\partial y} (y - x^{2})

Apply the Sum/Difference Rule:

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

= 20 (\frac{\partial}{\partial y} (y) - \frac{\partial}{\partial y} (x^{2}))

\frac{\partial}{\partial y} (y) = 1

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

= 20 (1 - 0)

Simplify

= 20

x \to 1 lim (\frac{x ^{2} + 1}{x + 5})

\int \frac{( x ^{3} - x )}{x} d x

(x + y)^{2} d x + (2 x y + x^{2} - 6) d y = 0, y (1) = 1

\frac{d}{d x} (\frac{- 6}{x})

\frac{d ^{2}}{d x ^{2}} (\frac{3 x + 5}{x ^{2} - x})

What is (\partial)/(\partial y)(20(y-x^2)) ?
The answer to (\partial)/(\partial y)(20(y-x^2)) is 20