What is the d/(dy)(4arctan(x/y)) ?

The d/(dy)(4arctan(x/y)) is -(4x)/(x^2+y^2)

What is the first d/(dy)(4arctan(x/y)) ?

The first d/(dy)(4arctan(x/y)) is -(4x)/(x^2+y^2)

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\frac{d}{d y} (4 arctan (\frac{x}{y}))

- \frac{4 x}{x ^{2} + y ^{2}}

Solution steps

\frac{d}{d y} (4 arctan (\frac{x}{y}))

Treat

x

as a constant

Take the constant out:

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

= 4 \frac{d}{d y} (arctan (\frac{x}{y}))

Apply the chain rule:

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

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

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

= 4 \cdot \frac{1}{( \frac{x}{y} ) ^{2} + 1} (- \frac{x}{y ^{2}})

Simplify

4 \cdot \frac{1}{( \frac{x}{y} ) ^{2} + 1} (- \frac{x}{y ^{2}}) : - \frac{4 x}{x ^{2} + y ^{2}}

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

\frac{\partial}{\partial x} (4 y^{2} + 7 x^{2})

x^{2} - 3, 1

\int sin^{4} (6 θ) d θ

x y^{^{'}} + 2 y = - 3 e^{x}

What is the d/(dy)(4arctan(x/y)) ?
The d/(dy)(4arctan(x/y)) is -(4x)/(x^2+y^2)
What is the first d/(dy)(4arctan(x/y)) ?
The first d/(dy)(4arctan(x/y)) is -(4x)/(x^2+y^2)