Find the Derivative - d/dz natural log of x+y^4-z

$\ln (x + y^{4}) - z$

Step 1

Differentiate.

By the Sum Rule, the derivative of $\ln (x + y^{4}) - z$ with respect to $z$ is $\frac{d}{d z} [\ln (x + y^{4})] + \frac{d}{d z} [- z]$ .

$\frac{d}{d z} [\ln (x + y^{4})] + \frac{d}{d z} [- z]$

Since $\ln (x + y^{4})$ is constant with respect to $z$ , the derivative of $\ln (x + y^{4})$ with respect to $z$ is $0$ .

$0 + \frac{d}{d z} [- z]$

Step 2

Evaluate $\frac{d}{d z} [- z]$ .

Since $- 1$ is constant with respect to $z$ , the derivative of $- z$ with respect to $z$ is $- \frac{d}{d z} [z]$ .

$0 - \frac{d}{d z} [z]$

Differentiate using the Power Rule which states that $\frac{d}{d z} [z^{n}]$ is $n z^{n - 1}$ where $n = 1$ .

$0 - 1 \cdot 1$

Multiply $- 1$ by $1$ .

$0 - 1$

Step 3

Subtract $1$ from $0$ .

$- 1$