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

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

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

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

2 x + 2 y

Solution steps

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

Simplify

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

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

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

\frac{\partial}{\partial y} (1) = 0

= 0 + 2 x + 0 + 2 y + 0

Simplify

= 2 x + 2 y

n = 2 \sum \infty \frac{1}{n 4 ^{n}}

6 e^{x} cos (x)

f (x) = - \frac{1}{( 5 x + 1 ) ^{2}}

4 y^{^{''}} + y^{^{'}} = 0

x \to 2 lim (\frac{2}{0})

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