Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

Step 2

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

$Q'$

Step 3

Differentiate the right side of the equation.

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Step 3.1

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

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

Step 3.2

Evaluate $\frac{d}{d L} [5 L]$ .

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Step 3.2.1

Since $5$ is constant with respect to $L$ , the derivative of $5 L$ with respect to $L$ is $5 \frac{d}{d L} [L]$ .

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

Step 3.2.2

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

$5 \cdot 1 + \frac{d}{d L} [6 L K^{2}] + \frac{d}{d L} [8 K]$

Step 3.2.3

Multiply $5$ by $1$ .

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

Step 3.3

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

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Step 3.3.1

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

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

Step 3.3.2

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

$5 + 6 K^{2} \cdot 1 + \frac{d}{d L} [8 K]$

Step 3.3.3

Multiply $6$ by $1$ .

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

Step 3.4

Since $8 K$ is constant with respect to $L$ , the derivative of $8 K$ with respect to $L$ is $0$ .

$5 + 6 K^{2} + 0$

Step 3.5

Simplify.

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Step 3.5.1

Add $5 + 6 K^{2}$ and $0$ .

$5 + 6 K^{2}$

Step 3.5.2

Reorder terms.

$6 K^{2} + 5$

Step 4

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

$Q' = 6 K^{2} + 5$

Step 5

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

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