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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int (x^{2} \cdot ln (x)) d x

\frac{1}{t ^{4}}

\int (sin (y) - y sin (x)) d x

\frac{\partial}{\partial t} (x + y sin (t))

\frac{d}{d x} (x^{2} + c^{2})

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