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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int \frac{6 x ^{3}}{x ^{2} + 9} d x

t \to - 1 lim (t^{2} - 1)

\int sin (5 x) \cdot cos (2 x) d x

f (t) = ln (t^{9} + 5)

y = 4 x - x^{2}, (1, 3)

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