What is (\partial)/(\partial x)(ln(x^2+6y^2+8)) ?

The answer to (\partial)/(\partial x)(ln(x^2+6y^2+8)) is (2x)/(x^2+6y^2+8)

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(\partial)/(\partial x)(ln(x^2+6y^2+8))

\frac{\partial}{\partial x} (ln (x^{2} + 6 y^{2} + 8))

\frac{2 x}{x ^{2} + 6 y ^{2} + 8}

Solution steps

\frac{\partial}{\partial x} (ln (x^{2} + 6 y^{2} + 8))

Treat

y

as a constant

Apply the chain rule:

\frac{1}{x ^{2} + 6 y ^{2} + 8} \frac{\partial}{\partial x} (x^{2} + 6 y^{2} + 8)

= \frac{1}{x ^{2} + 6 y ^{2} + 8} \frac{\partial}{\partial x} (x^{2} + 6 y^{2} + 8)

\frac{\partial}{\partial x} (x^{2} + 6 y^{2} + 8) = 2 x

= \frac{1}{x ^{2} + 6 y ^{2} + 8} \cdot 2 x

Simplify

\frac{1}{x ^{2} + 6 y ^{2} + 8} \cdot 2 x : \frac{2 x}{x ^{2} + 6 y ^{2} + 8}

= \frac{2 x}{x ^{2} + 6 y ^{2} + 8}

\frac{d y}{d x} = (y + 8)

\int \frac{sec ( θ )}{tan ( θ )} d θ

\frac{d y}{d x} + x \cdot y = - x^{2}

\frac{d}{d x} (x (x - 4)^{2})

2 x^{2} + 3 x, at x = - 3

What is (\partial)/(\partial x)(ln(x^2+6y^2+8)) ?
The answer to (\partial)/(\partial x)(ln(x^2+6y^2+8)) is (2x)/(x^2+6y^2+8)