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

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

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

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

\frac{1}{2 x + y + 1}

Solution steps

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

Treat

x

as a constant

Apply the chain rule:

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

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

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

= \frac{1}{2 x + y + 1} \cdot 1

Simplify

= \frac{1}{2 x + y + 1}

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