What is (\partial)/(\partial x)((1-y)+10(x-y^2)^2) ?

The answer to (\partial)/(\partial x)((1-y)+10(x-y^2)^2) is 20(x-y^2)

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(\partial)/(\partial x)((1-y)+10(x-y^2)^2)

\frac{\partial}{\partial x} ((1 - y) + 10 (x - y^{2})^{2})

20 (x - y^{2})

Solution steps

\frac{\partial}{\partial x} ((1 - y) + 10 (x - y^{2})^{2})

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (1) - \frac{\partial}{\partial x} (y) + \frac{\partial}{\partial x} (10 (x - y^{2})^{2})

\frac{\partial}{\partial x} (1) = 0

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

\frac{\partial}{\partial x} (10 (x - y^{2})^{2}) = 20 (x - y^{2})

= 0 - 0 + 20 (x - y^{2})

Simplify

= 20 (x - y^{2})

x y^{^{'}} + 3 y = x^{2} - 1

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

(x + a) e^{3 x} + b

2 x \frac{d y}{d x} = 1 - y^{2}

\int_{0}^{1} (e^{x}) d x

What is (\partial)/(\partial x)((1-y)+10(x-y^2)^2) ?
The answer to (\partial)/(\partial x)((1-y)+10(x-y^2)^2) is 20(x-y^2)