What is (\partial)/(\partial x)(ln(5x^2-2y^2)) ?

The answer to (\partial)/(\partial x)(ln(5x^2-2y^2)) is (10x)/(5x^2-2y^2)

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(\partial)/(\partial x)(ln(5x^2-2y^2))

\frac{\partial}{\partial x} (ln (5 x^{2} - 2 y^{2}))

\frac{10 x}{5 x ^{2} - 2 y ^{2}}

Solution steps

\frac{\partial}{\partial x} (ln (5 x^{2} - 2 y^{2}))

Treat

y

as a constant

Apply the chain rule:

\frac{1}{5 x ^{2} - 2 y ^{2}} \frac{\partial}{\partial x} (5 x^{2} - 2 y^{2})

= \frac{1}{5 x ^{2} - 2 y ^{2}} \frac{\partial}{\partial x} (5 x^{2} - 2 y^{2})

\frac{\partial}{\partial x} (5 x^{2} - 2 y^{2}) = 10 x

= \frac{1}{5 x ^{2} - 2 y ^{2}} \cdot 10 x

Simplify

\frac{1}{5 x ^{2} - 2 y ^{2}} \cdot 10 x : \frac{10 x}{5 x ^{2} - 2 y ^{2}}

= \frac{10 x}{5 x ^{2} - 2 y ^{2}}

\int_{1}^{3} \frac{2}{4 + ( x - 1 ) ^{2}} d x

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

x \to 0 lim (\frac{x ^{2}}{x})

\frac{d}{d x} (2 sec (cos (x)) + 2 tan (2 x))

x \to 5 lim (\frac{x ^{2} + k x - 20}{x - 5})

What is (\partial)/(\partial x)(ln(5x^2-2y^2)) ?
The answer to (\partial)/(\partial x)(ln(5x^2-2y^2)) is (10x)/(5x^2-2y^2)