What is (\partial)/(\partial x)(ln(1+x^2y^2)) ?

The answer to (\partial)/(\partial x)(ln(1+x^2y^2)) is (2xy^2)/(1+x^2y^2)

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(\partial)/(\partial x)(ln(1+x^2y^2))

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

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

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d y}{d x} = \frac{x - 2}{y - 2}

\int 10 x^{9} + 10 x^{5} - 2 x^{4} - 1 d x

\frac{1}{x ^{8}}

sin (x) ln (sec (x) + tan (x)) - 1

x^{- 2}, 7

What is (\partial)/(\partial x)(ln(1+x^2y^2)) ?
The answer to (\partial)/(\partial x)(ln(1+x^2y^2)) is (2xy^2)/(1+x^2y^2)