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

The answer to (\partial)/(\partial x)(ln(x-y^2)) is 1/(x-y^2)

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

\frac{d y}{d x} = \frac{4 sec ( y )}{( x - 6 ) ^{2}}

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

\frac{d x}{d y} = 4 (x^{2} + 1)

x \to \infty lim (\frac{4 x ^{3} + 8}{4 x ^{3} + 64 x ^{6} + 3})

\frac{d}{d x} (\frac{1 - 2 x}{1 + 2 x})

What is (\partial)/(\partial x)(ln(x-y^2)) ?
The answer to (\partial)/(\partial x)(ln(x-y^2)) is 1/(x-y^2)