What is (\partial)/(\partial y)(zarctan(y/x)) ?

The answer to (\partial)/(\partial y)(zarctan(y/x)) is (zx)/(y^2+x^2)

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(\partial)/(\partial y)(zarctan(y/x))

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

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

Solution steps

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

Treat

z, x

as constants

Take the constant out:

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

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

Apply the chain rule:

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

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

\frac{\partial}{\partial y} (\frac{y}{x}) = \frac{1}{x}

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

Simplify

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

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

f (x) = \frac{1}{x ln ( x )}

\int_{0}^{1} 64 t^{2} + 9 t^{4} d t

tan (3 t)

\int ln (7 x + 5) d x

\int 2 t 2 t - 3 d t

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