Calculus Examples

$y^{x y}$

Step 1

Differentiate using the Generalized Power Rule which states that $\frac{d}{d x} [f^{g}]$ is $f^{g} (f' \frac{g}{f} + g' \ln (f))$ where $f = y$ and $g = x y$ .

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

Step 2

Multiply $y$ by $y^{x y - 1}$ by adding the exponents.

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Step 2.1

Move $y^{x y - 1}$ .

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

Step 2.2

Multiply $y^{x y - 1}$ by $y$ .

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Step 2.2.1

Raise $y$ to the power of $1$ .

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

Step 2.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 2.3

Combine the opposite terms in $x y - 1 + 1$ .

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Step 2.3.1

Add $- 1$ and $1$ .

$\frac{d}{d x} [y] (x y^{x y + 0}) + \frac{d}{d x} [x y] (y^{x y} \ln (y))$

Step 2.3.2

Add $x y$ and $0$ .

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

Step 3

Since $y$ is constant with respect to $x$ , the derivative of $y$ with respect to $x$ is $0$ .

$0 (x y^{x y}) + \frac{d}{d x} [x y] (y^{x y} \ln (y))$

Step 4

Simplify the expression.

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Step 4.1

Multiply $x$ by $0$ .

$0 y^{x y} + \frac{d}{d x} [x y] (y^{x y} \ln (y))$

Step 4.2

Multiply $0$ by $y^{x y}$ .

$0 + \frac{d}{d x} [x y] (y^{x y} \ln (y))$

Step 4.3

Add $0$ and $\frac{d}{d x} [x y] (y^{x y} \ln (y))$ .

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

Step 5

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

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

Step 6

Raise $y$ to the power of $1$ .

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

Step 7

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 8

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

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

Step 9

Multiply $y^{x y + 1}$ by $1$ .

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