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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

y^{^{'}} t^{4} = 0

\int_{0}^{π} x^{2} sin (x) d x

e^{- 4 t} \cdot cos (t)

\int (- \frac{2}{x ^{2}} + 3 x) d x

\int \frac{1}{3 - x ^{2} - 2 x} d x

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