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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d}{d y} (\frac{y ^{3}}{8})

t an g e n t y = x^{4} + 8 x^{2} - x, (1, 8)

\int x + 2 - \frac{1}{x + 1} d x

\int \frac{x + 1}{x ^{2} + 2 x} d x

n = 1 \sum \infty \frac{1}{3} (\frac{e}{2})^{n}

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