What is (\partial)/(\partial x)((x+2)^2-2(y-1)^2-5) ?

The answer to (\partial)/(\partial x)((x+2)^2-2(y-1)^2-5) is 2x+4

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(\partial)/(\partial x)((x+2)^2-2(y-1)^2-5)

\frac{\partial}{\partial x} ((x + 2)^{2} - 2 (y - 1)^{2} - 5)

2 x + 4

Solution steps

\frac{\partial}{\partial x} ((x + 2)^{2} - 2 (y - 1)^{2} - 5)

Simplify

(x + 2)^{2} - 2 (y - 1)^{2} - 5 : x^{2} + 4 x + 4 y - 2 y^{2} - 3

= \frac{\partial}{\partial x} (x^{2} + 4 x + 4 y - 2 y^{2} - 3)

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} (4 x) + \frac{\partial}{\partial x} (4 y) - \frac{\partial}{\partial x} (2 y^{2}) - \frac{\partial}{\partial x} (3)

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

\frac{\partial}{\partial x} (4 x) = 4

\frac{\partial}{\partial x} (4 y) = 0

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

\frac{\partial}{\partial x} (3) = 0

= 2 x + 4 + 0 - 0 - 0

Simplify

= 2 x + 4

sin (x) + cos (x)

y^{^{''}} + 4 y^{^{'}} + 3 y = x

f (x) = x + 5, at x = 11

3 x - 5

\int \frac{1}{x} e^{- x} d x

What is (\partial)/(\partial x)((x+2)^2-2(y-1)^2-5) ?
The answer to (\partial)/(\partial x)((x+2)^2-2(y-1)^2-5) is 2x+4