What is (\partial)/(\partial y)(ln(2xy+4)) ?

The answer to (\partial)/(\partial y)(ln(2xy+4)) is x/(xy+2)

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(\partial)/(\partial y)(ln(2xy+4))

\frac{\partial}{\partial y} (ln (2 x y + 4))

\frac{x}{x y + 2}

Solution steps

\frac{\partial}{\partial y} (ln (2 x y + 4))

Treat

x

as a constant

Apply the chain rule:

\frac{1}{2 x y + 4} \frac{\partial}{\partial y} (2 x y + 4)

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

\frac{\partial}{\partial y} (2 x y + 4) = 2 x

= \frac{1}{2 x y + 4} \cdot 2 x

Simplify

\frac{1}{2 x y + 4} \cdot 2 x : \frac{x}{x y + 2}

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

f (x) = x^{c o s (x)}

f (x) = x x, at x = 18

x \to 2 + lim (\frac{1 - x}{x - 2})

\int_{0}^{2} \frac{1}{4 - x ^{2}} d x

\int \frac{5 x ^{2} + 15 x + 8}{( x + 2 ) ^{2} x} d x

What is (\partial)/(\partial y)(ln(2xy+4)) ?
The answer to (\partial)/(\partial y)(ln(2xy+4)) is x/(xy+2)