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

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

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

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

2 x + 2

Solution steps

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

Simplify

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

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

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

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

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

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

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

= 2 x + 2 + 0 + 0 - 0

Simplify

= 2 x + 2

x^{2} y^{^{'}} - 3 x y = 4 x^{6}

\int 16 - x d x

\int \frac{x ^{5}}{x ^{6} - 4} d x

\frac{d}{d x} (csc (x) tan (x))

4 sin (t) \frac{d y}{d t} + 4 y cos (t) = cos (t) sin^{4} (t)

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