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

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

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

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

- 1

Solution steps

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

Treat

x, y

as constants

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial z} (ln (1 + x y)) = 0

\frac{\partial}{\partial z} (z) = 1

= 0 - 1

Simplify

= - 1

f (x) = x + 8

\int t ln (1 + t) d t

\int_{- 8}^{0} (\frac{y}{8} + y + 9) d y

x \to \infty lim (x + 9)

\int \frac{1}{x ^{2} + 3 x + 3} d x

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