What is (\partial)/(\partial y)(ln(3+x^2y^2)) ?

The answer to (\partial)/(\partial y)(ln(3+x^2y^2)) is (2x^2y)/(3+x^2y^2)

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(\partial)/(\partial y)(ln(3+x^2y^2))

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

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

Solution steps

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

Treat

x

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d}{d x} (\frac{sin ( x )}{1 + 2 cos ( x )})

y = e^{x} cos (x)

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

\frac{d}{d x} (sin (x) + x)

\int \frac{2 x}{1 - 4 x ^{4}} d x

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