What is (\partial}{\partial x}(zarctan(\frac{3y)/x)) ?

The answer to (\partial}{\partial x}(zarctan(\frac{3y)/x)) is -(3zy)/(9y^2+x^2)

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(\partial}{\partial x}(zarctan(\frac{3y)/x))

\frac{\partial}{\partial x} (z arctan (\frac{3 y}{x}))

- \frac{3 zy}{9 y ^{2} + x ^{2}}

Solution steps

\frac{\partial}{\partial x} (z arctan (\frac{3 y}{x}))

Treat

z, y

as constants

Take the constant out:

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

= z \frac{\partial}{\partial x} (arctan (\frac{3 y}{x}))

Apply the chain rule:

\frac{1}{( \frac{3 y}{x} ) ^{2} + 1} \frac{\partial}{\partial x} (\frac{3 y}{x})

= \frac{1}{( \frac{3 y}{x} ) ^{2} + 1} \frac{\partial}{\partial x} (\frac{3 y}{x})

\frac{\partial}{\partial x} (\frac{3 y}{x}) = - \frac{3 y}{x ^{2}}

= z \frac{1}{( \frac{3 y}{x} ) ^{2} + 1} (- \frac{3 y}{x ^{2}})

Simplify

z \frac{1}{( \frac{3 y}{x} ) ^{2} + 1} (- \frac{3 y}{x ^{2}}) : - \frac{3 zy}{9 y ^{2} + x ^{2}}

= - \frac{3 zy}{9 y ^{2} + x ^{2}}

\int_{- 2.5}^{0} e^{3 x} d x

\int x^{3} \cdot e^{x} d x

\frac{\partial}{\partial x} (2 x y + x^{3})

\frac{d}{d x} (1 - x^{4})

\int e^{3 x + 5} \cdot (9 x + 5) d x

What is (\partial}{\partial x}(zarctan(\frac{3y)/x)) ?
The answer to (\partial}{\partial x}(zarctan(\frac{3y)/x)) is -(3zy)/(9y^2+x^2)