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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int - \frac{1}{4} d x

\int t \cdot sin (t^{2}) d t

\frac{d}{d x} (- sin (x - 2))

n = 0 \sum \infty - 6 \cdot (- \frac{1}{3})^{n}

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