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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

n = 0 \sum \infty \frac{2 n}{2 ^{n}}

\int \frac{d}{1} (e^{x^{2}} y) d x

\frac{d y}{d x} = - 5 + x + y

\frac{200}{0.9 + 0.1 e ^{2 h}}

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