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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{x ^{5}}{4 - x ^{4}}

y^{^{'''}} - 7 y^{^{''}} - 8 y^{^{'}} = 0

f (x) = \frac{2}{x ^{2}}, at x = 1

\frac{d}{d x} (\frac{1}{2} cos^{2} (\frac{x}{2}))

\int x^{3} \cdot cos (x^{2}) d x

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