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

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

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

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

- \frac{2 y}{y ^{2} + x ^{2}}

Solution steps

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

Treat

y

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the chain rule:

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

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

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

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

Simplify

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

= - \frac{2 y}{y ^{2} + x ^{2}}

y = 7 sec (x), at x = \frac{π}{3}

\frac{d}{d t} (arctan (\frac{t}{2}))

x^{2} + 5

y = \frac{sec ( x )}{1 + sec ( x )}

\int 19 sin^{- \frac{3}{2}} (x) cos^{3} (x) d x

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