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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

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

\int_{1}^{3} 4 x^{3} - 8 x + 1 d x

9 t sin (π t)

\frac{1}{( 1 - x ) ^{3}}

\int - 2 x^{2} + 4 d x

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