What is (\partial)/(\partial x)(-(x-4)^2-(y-1)^2) ?

The answer to (\partial)/(\partial x)(-(x-4)^2-(y-1)^2) is -2x+8

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(\partial)/(\partial x)(-(x-4)^2-(y-1)^2)

\frac{\partial}{\partial x} (- (x - 4)^{2} - (y - 1)^{2})

- 2 x + 8

Solution steps

\frac{\partial}{\partial x} (- (x - 4)^{2} - (y - 1)^{2})

Simplify

- (x - 4)^{2} - (y - 1)^{2} : - x^{2} + 8 x + 2 y - y^{2} - 17

= \frac{\partial}{\partial x} (- x^{2} + 8 x + 2 y - y^{2} - 17)

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} (2 y) - \frac{\partial}{\partial x} (y^{2}) - \frac{\partial}{\partial x} (17)

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

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

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

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

\frac{\partial}{\partial x} (17) = 0

= - 2 x + 8 + 0 - 0 - 0

Simplify

= - 2 x + 8

\int \frac{1}{x} d x

\int \frac{1}{81 + 16 x ^{2}} d x

y^{^{'}} + y = sin (2 t)

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

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

What is (\partial)/(\partial x)(-(x-4)^2-(y-1)^2) ?
The answer to (\partial)/(\partial x)(-(x-4)^2-(y-1)^2) is -2x+8