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

The answer to (\partial)/(\partial x)((1-x)+10(y-x^2)^2) is -40x(y-x^2)-1

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

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

- 40 x (y - x^{2}) - 1

Solution steps

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

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

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

\frac{\partial}{\partial x} (10 (y - x^{2})^{2}) = - 40 x (y - x^{2})

= 0 - 1 - 40 x (y - x^{2})

Simplify

= - 40 x (y - x^{2}) - 1

\int_{1}^{\infty} \frac{1}{1 + x ^{2}} d x

\frac{d}{d x} (x - 2 - 0.5 sin (x))

\frac{d}{d s} (s t)

f (x) = - 4 sin (x)

2 x + 5

What is (\partial)/(\partial x)((1-x)+10(y-x^2)^2) ?
The answer to (\partial)/(\partial x)((1-x)+10(y-x^2)^2) is -40x(y-x^2)-1