What is (\partial)/(\partial x)(ln(x^2+3y^2+8)) ?

The answer to (\partial)/(\partial x)(ln(x^2+3y^2+8)) is (2x)/(x^2+3y^2+8)

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(\partial)/(\partial x)(ln(x^2+3y^2+8))

\frac{\partial}{\partial x} (ln (x^{2} + 3 y^{2} + 8))

\frac{2 x}{x ^{2} + 3 y ^{2} + 8}

Solution steps

\frac{\partial}{\partial x} (ln (x^{2} + 3 y^{2} + 8))

Treat

y

as a constant

Apply the chain rule:

\frac{1}{x ^{2} + 3 y ^{2} + 8} \frac{\partial}{\partial x} (x^{2} + 3 y^{2} + 8)

= \frac{1}{x ^{2} + 3 y ^{2} + 8} \frac{\partial}{\partial x} (x^{2} + 3 y^{2} + 8)

\frac{\partial}{\partial x} (x^{2} + 3 y^{2} + 8) = 2 x

= \frac{1}{x ^{2} + 3 y ^{2} + 8} \cdot 2 x

Simplify

\frac{1}{x ^{2} + 3 y ^{2} + 8} \cdot 2 x : \frac{2 x}{x ^{2} + 3 y ^{2} + 8}

= \frac{2 x}{x ^{2} + 3 y ^{2} + 8}

\int \frac{1}{( x - 2 ) ( x - 4 ) ( x + 1 )} d x

\frac{d}{d x} (a x^{4} - 2 a x^{3} + 2 b x^{2} \cdot ln (\frac{x}{2}))

x + 2

x = - y^{2} + 4 y, [1, 4]

x \to 0 lim (sin^{2} (\frac{π}{x}))

What is (\partial)/(\partial x)(ln(x^2+3y^2+8)) ?
The answer to (\partial)/(\partial x)(ln(x^2+3y^2+8)) is (2x)/(x^2+3y^2+8)