Solve yz-\ln{z}=x+y

Evaluate

yz - ln (z) = x + y

Find \frac{\partial z}{\partial x} by taking the derivative of both sides with respect to x

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

Use differentiation rule \frac{\partial}{\partial x} (f (x) \pm g (x)) = \frac{\partial}{\partial x} (f (x)) \pm \frac{\partial}{\partial x} (g (x))

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

Evaluate

More Steps

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

Evaluate

More Steps

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

Reduce fractions to a common denominator

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

Write all numerators above the common denominator

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

Use the commutative property to reorder the terms

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

Use differentiation rule \frac{\partial}{\partial x} (f (x) \pm g (x)) = \frac{\partial}{\partial x} (f (x)) \pm \frac{\partial}{\partial x} (g (x))

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

Use \frac{\partial}{\partial x} x^{n} = n x^{n - 1} to find derivative

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

Use \frac{\partial}{\partial x} (c) = 0 to find derivative

\frac{yz \frac{\partial z}{\partial x} - \frac{\partial z}{\partial x}}{z} = 1 + 0

Removing 0 doesn't change the value,so remove it from the expression

\frac{yz \frac{\partial z}{\partial x} - \frac{\partial z}{\partial x}}{z} = 1

Cross multiply

yz \frac{\partial z}{\partial x} - \frac{\partial z}{\partial x} = z

Collect like terms by calculating the sum or difference of their coefficients

(yz - 1) \frac{\partial z}{\partial x} = z

Divide both sides

\frac{( yz - 1 ) \frac{\partial z}{\partial x}}{yz - 1} = \frac{z}{yz - 1}

Solution

\frac{\partial z}{\partial x} = \frac{z}{yz - 1}