What is (\partial)/(\partial x)((x-4)^2+(y-4)^2-5) ?

The answer to (\partial)/(\partial x)((x-4)^2+(y-4)^2-5) is 2x-8

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(\partial)/(\partial x)((x-4)^2+(y-4)^2-5)

\frac{\partial}{\partial x} ((x - 4)^{2} + (y - 4)^{2} - 5)

2 x - 8

Solution steps

\frac{\partial}{\partial x} ((x - 4)^{2} + (y - 4)^{2} - 5)

Simplify

(x - 4)^{2} + (y - 4)^{2} - 5 : x^{2} - 8 x + y^{2} + 27 - 8 y

= \frac{\partial}{\partial x} (x^{2} - 8 x + y^{2} + 27 - 8 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} (8 x) + \frac{\partial}{\partial x} (y^{2}) + \frac{\partial}{\partial x} (27) - \frac{\partial}{\partial x} (8 y)

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

\frac{\partial}{\partial x} (8 x) = 8

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

\frac{\partial}{\partial x} (27) = 0

\frac{\partial}{\partial x} (8 y) = 0

= 2 x - 8 + 0 + 0 - 0

Simplify

= 2 x - 8

\int \frac{x}{16 - 9 x ^{4}} d x

\int \frac{3 x ^{2} - 10}{x ^{2} - 4 x + 4} d x

y^{^{''}} - 6 y^{^{'}} + 13 y = 0

\frac{d y}{d x} = x y^{3} (1 + x^{2})^{- \frac{1}{2}}, y (0) = 1

\frac{\partial}{\partial x} (- x y (- 15 - x - y))

What is (\partial)/(\partial x)((x-4)^2+(y-4)^2-5) ?
The answer to (\partial)/(\partial x)((x-4)^2+(y-4)^2-5) is 2x-8