Find dz/dy z=x^y+ natural log of xy

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

Step 1

Differentiate both sides of the equation.

$\frac{d}{d y} (z) = \frac{d}{d y} (x^{y} + \ln (x y))$

Step 2

The derivative of $z$ with respect to $y$ is $z'$ .

$z'$

Step 3

Differentiate the right side of the equation.

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

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

Differentiate using the Exponential Rule which states that $\frac{d}{d y} [a^{y}]$ is $a^{y} \ln (a)$ where $a$ = $x$ .

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

Evaluate $\frac{d}{d y} [\ln (x y)]$ .

Differentiate using the chain rule, which states that $\frac{d}{d y} [f (g (y))]$ is $f' (g (y)) g' (y)$ where $f (y) = \ln (y)$ and $g (y) = x y$ .

To apply the Chain Rule, set $u$ as $x y$ .

$x^{y} \ln (x) + \frac{d}{d u} [\ln (u)] \frac{d}{d y} [x y]$

The derivative of $\ln (u)$ with respect to $u$ is $\frac{1}{u}$ .

$x^{y} \ln (x) + \frac{1}{u} \frac{d}{d y} [x y]$

Replace all occurrences of $u$ with $x y$ .

$x^{y} \ln (x) + \frac{1}{x y} \frac{d}{d y} [x y]$

Since $x$ is constant with respect to $y$ , the derivative of $x y$ with respect to $y$ is $x \frac{d}{d y} [y]$ .

$x^{y} \ln (x) + \frac{1}{x y} (x \frac{d}{d y} [y])$

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

$x^{y} \ln (x) + \frac{1}{x y} (x \cdot 1)$

Multiply $x$ by $1$ .

$x^{y} \ln (x) + \frac{1}{x y} x$

Combine $\frac{1}{x y}$ and $x$ .

$x^{y} \ln (x) + \frac{x}{x y}$

Cancel the common factor of $x$ .

Cancel the common factor.

$x^{y} \ln (x) + \frac{x}{x y}$

Rewrite the expression.

$x^{y} \ln (x) + \frac{1}{y}$

Step 4

Reform the equation by setting the left side equal to the right side.

$z' = x^{y} \ln (x) + \frac{1}{y}$

Step 5

Replace $z'$ with $\frac{d z}{d y}$ .

$\frac{d z}{d y} = x^{y} \ln (x) + \frac{1}{y}$