What is (\partial)/(\partial y)(ln(1+xy)-z) ?

The answer to (\partial)/(\partial y)(ln(1+xy)-z) is x/(xy+1)

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(\partial)/(\partial y)(ln(1+xy)-z)

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

\frac{x}{x y + 1}

Solution steps

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

Treat

x, z

as constants

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{\partial}{\partial y} (ln (1 + x y)) - \frac{\partial}{\partial y} (z)

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

\frac{\partial}{\partial y} (z) = 0

= \frac{x}{1 + x y} - 0

Simplify

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

\frac{\partial}{\partial θ} (ρ cos (φ))

x \to 5 lim (2 x - 3 x + 4)

\int \frac{y ^{2} + 38 y + 361}{( y ^{2} + 361 ) ^{2}} d y

x \to 0 - lim (8 + \frac{7}{x})

\int \frac{3 r ^{2}}{r + 9} d r

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