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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

\int \frac{18}{( x ^{2} - 1 ) ^{2}} d x

y^{^{'}} + 7 x^{- 1} y = x^{2}

10 arcsin (e^{\frac{x}{2}})

\int \frac{4 x ^{2}}{x ^{2} + 7} d x

x \to 3 lim (\frac{1}{x - 2})

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