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

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

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

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

2 x - 2

Solution steps

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

Simplify

(x - 1)^{2} + (y - 1)^{2} : x^{2} - 2 x + y^{2} + 2 - 2 y

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

\frac{\partial}{\partial x} (2 x) = 2

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

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

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

= 2 x - 2 + 0 + 0 - 0

Simplify

= 2 x - 2

\int_{- 1}^{1} \frac{tan ( x )}{1 + x ^{4}} d x

\int_{- \infty}^{0} \frac{1}{2 - 8 x} d x

x = y^{4}, 2 - x, y = 0

\frac{4 x - 6}{2 x ^{2} + x - 2}

\int \frac{8 x ^{2}}{25 - x ^{2}} d x

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