What is (\partial)/(\partial y)(arctan(y^3x)) ?

The answer to (\partial)/(\partial y)(arctan(y^3x)) is (3y^2x)/(y^6x^2+1)

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(\partial)/(\partial y)(arctan(y^3x))

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

\frac{3 y ^{2} x}{y ^{6} x ^{2} + 1}

Solution steps

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

Treat

x

as a constant

Apply the chain rule:

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

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

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

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

Simplify

\frac{1}{( y ^{3} x ) ^{2} + 1} \cdot 3 x y^{2} : \frac{3 y ^{2} x}{y ^{6} x ^{2} + 1}

= \frac{3 y ^{2} x}{y ^{6} x ^{2} + 1}

\int \frac{1}{1 + x ^{2}} d x

y = sin (arccos (t))

\frac{d}{d x} (\frac{15}{x ^{2}})

\int_{0}^{12} \frac{1}{36 - x ^{2}} d x

\int_{0}^{\frac{π}{6}} t sin (2 t) d t

What is (\partial)/(\partial y)(arctan(y^3x)) ?
The answer to (\partial)/(\partial y)(arctan(y^3x)) is (3y^2x)/(y^6x^2+1)