What is (\partial)/(\partial x)(x^2+y^2+(21-x-y)^2) ?

The answer to (\partial)/(\partial x)(x^2+y^2+(21-x-y)^2) is 4x+2y-42

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(\partial)/(\partial x)(x^2+y^2+(21-x-y)^2)

\frac{\partial}{\partial x} (x^{2} + y^{2} + (21 - x - y)^{2})

4 x + 2 y - 42

Solution steps

\frac{\partial}{\partial x} (x^{2} + y^{2} + (21 - x - y)^{2})

Simplify

x^{2} + y^{2} + (21 - x - y)^{2} : 2 x^{2} + 2 x y - 42 x + 2 y^{2} - 42 y + 441

= \frac{\partial}{\partial x} (2 x^{2} + 2 x y - 42 x + 2 y^{2} - 42 y + 441)

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (2 x^{2}) + \frac{\partial}{\partial x} (2 x y) - \frac{\partial}{\partial x} (42 x) + \frac{\partial}{\partial x} (2 y^{2}) - \frac{\partial}{\partial x} (42 y) + \frac{\partial}{\partial x} (441)

\frac{\partial}{\partial x} (2 x^{2}) = 4 x

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

\frac{\partial}{\partial x} (42 x) = 42

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

\frac{\partial}{\partial x} (42 y) = 0

\frac{\partial}{\partial x} (441) = 0

= 4 x + 2 y - 42 + 0 - 0 + 0

Simplify

= 4 x + 2 y - 42

\frac{d ^{2} y}{d t ^{2}} - \frac{d y}{d t} + 225 y = 15 sin (15 t)

\int - 2 sin (x) cos (x) d x

\frac{\partial}{\partial x} (5 x^{2} y + 2 x y^{3} + 3 y^{2})

x^{^{''}} + x = 0, x (0) = 0, x^{^{'}} (0) = 0

\int_{- 4}^{4} (16 - x^{2}) d x

What is (\partial)/(\partial x)(x^2+y^2+(21-x-y)^2) ?
The answer to (\partial)/(\partial x)(x^2+y^2+(21-x-y)^2) is 4x+2y-42