What is (\partial)/(\partial y)(ln(y)ln(x^2+1)) ?

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

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(\partial)/(\partial y)(ln(y)ln(x^2+1))

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

\frac{ln ( x ^{2} + 1 )}{y}

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply the common derivative:

\frac{\partial}{\partial y} (ln (y)) = \frac{1}{y}

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

Simplify

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

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

\int (\frac{x ^{3}}{3})^{'} sin (x) d x

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

1 + sec (2 x)

cos (θ)

x \to 0 lim (e^{x})

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