Calculus Examples

$Q = 5 L + 6 L K^{2} + 8 K$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d K} (Q) = \frac{d}{d K} (5 L + 6 L K^{2} + 8 K)$

Step 2

The derivative of $Q$ with respect to $K$ is $Q'$ .

$Q'$

Step 3

Differentiate the right side of the equation.

Tap for more steps...

Step 3.1

Differentiate.

Tap for more steps...

Step 3.1.1

By the Sum Rule, the derivative of $5 L + 6 L K^{2} + 8 K$ with respect to $K$ is $\frac{d}{d K} [5 L] + \frac{d}{d K} [6 L K^{2}] + \frac{d}{d K} [8 K]$ .

$\frac{d}{d K} [5 L] + \frac{d}{d K} [6 L K^{2}] + \frac{d}{d K} [8 K]$

Step 3.1.2

Since $5 L$ is constant with respect to $K$ , the derivative of $5 L$ with respect to $K$ is $0$ .

$0 + \frac{d}{d K} [6 L K^{2}] + \frac{d}{d K} [8 K]$

Step 3.2

Evaluate $\frac{d}{d K} [6 L K^{2}]$ .

Tap for more steps...

Step 3.2.1

Since $6 L$ is constant with respect to $K$ , the derivative of $6 L K^{2}$ with respect to $K$ is $6 L \frac{d}{d K} [K^{2}]$ .

$0 + 6 L \frac{d}{d K} [K^{2}] + \frac{d}{d K} [8 K]$

Step 3.2.2

Differentiate using the Power Rule which states that $\frac{d}{d K} [K^{n}]$ is $n K^{n - 1}$ where $n = 2$ .

$0 + 6 L (2 K) + \frac{d}{d K} [8 K]$

Step 3.2.3

Multiply $2$ by $6$ .

$0 + 12 L K + \frac{d}{d K} [8 K]$

Step 3.3

Evaluate $\frac{d}{d K} [8 K]$ .

Tap for more steps...

Step 3.3.1

Since $8$ is constant with respect to $K$ , the derivative of $8 K$ with respect to $K$ is $8 \frac{d}{d K} [K]$ .

$0 + 12 L K + 8 \frac{d}{d K} [K]$

Step 3.3.2

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

$0 + 12 L K + 8 \cdot 1$

Step 3.3.3

Multiply $8$ by $1$ .

$0 + 12 L K + 8$

Step 3.4

Simplify.

Tap for more steps...

Step 3.4.1

Add $0$ and $12 L K$ .

$12 L K + 8$

Step 3.4.2

Reorder terms.

$12 K L + 8$

Step 4

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

$Q' = 12 K L + 8$

Step 5

Replace $Q'$ with $\frac{d Q}{d K}$ .

$\frac{d Q}{d K} = 12 K L + 8$