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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

t \cdot \frac{d y}{d t} + 14 y = \frac{1}{t}

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

\int \frac{( ln ( x ) ) ^{2}}{x} d x

(3 x^{2} - 1)^{4}

\int_{- 1}^{3} (x^{3} - 2 x) d x

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