What is (\partial)/(\partial y)(ln(10x^2-7y^2)) ?

The answer to (\partial)/(\partial y)(ln(10x^2-7y^2)) is -(14y)/(10x^2-7y^2)

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(\partial)/(\partial y)(ln(10x^2-7y^2))

\frac{\partial}{\partial y} (ln (10 x^{2} - 7 y^{2}))

- \frac{14 y}{10 x ^{2} - 7 y ^{2}}

Solution steps

\frac{\partial}{\partial y} (ln (10 x^{2} - 7 y^{2}))

Treat

x

as a constant

Apply the chain rule:

\frac{1}{10 x ^{2} - 7 y ^{2}} \frac{\partial}{\partial y} (10 x^{2} - 7 y^{2})

= \frac{1}{10 x ^{2} - 7 y ^{2}} \frac{\partial}{\partial y} (10 x^{2} - 7 y^{2})

\frac{\partial}{\partial y} (10 x^{2} - 7 y^{2}) = - 14 y

= \frac{1}{10 x ^{2} - 7 y ^{2}} (- 14 y)

Simplify

\frac{1}{10 x ^{2} - 7 y ^{2}} (- 14 y) : - \frac{14 y}{10 x ^{2} - 7 y ^{2}}

= - \frac{14 y}{10 x ^{2} - 7 y ^{2}}

f (x) = \frac{4 - 3 x - x ^{2}}{x ^{2} - 1}

\int 10 arcsin (x) d x

x \to - 2 lim ((x + 1) (x + 3))

x^{(\frac{2}{x})}

What is (\partial)/(\partial y)(ln(10x^2-7y^2)) ?
The answer to (\partial)/(\partial y)(ln(10x^2-7y^2)) is -(14y)/(10x^2-7y^2)