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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int t^{3} e^{- t} d t

f (x) = (x + 2)^{2} e^{- x}

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

sin (\frac{10 π}{x})

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