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

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

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

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

2 (x - 4)

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

= 2 (x - 4) + 0

Simplify

= 2 (x - 4)

(2 x + y cos (x y)) d x + x cos (x y) d y = 0

\frac{\partial}{\partial x} (x^{9} y^{4} - x^{8} y^{3})

x y^{^{'}} = y + x sin (x)

y^{^{''}} + 2 y^{^{'}} + 5 y = 0

\int \frac{1}{( x + 4 ) ^{3}} d x

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