Calculus Examples

$p = 400 e^{- 0.2 q}$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d q} (p) = \frac{d}{d q} (400 e^{- 0.2 q})$

Step 2

The derivative of $p$ with respect to $q$ is $p'$ .

$p'$

Step 3

Differentiate the right side of the equation.

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Step 3.1

Since $400$ is constant with respect to $q$ , the derivative of $400 e^{- 0.2 q}$ with respect to $q$ is $400 \frac{d}{d q} [e^{- 0.2 q}]$ .

$400 \frac{d}{d q} [e^{- 0.2 q}]$

Step 3.2

Differentiate using the chain rule, which states that $\frac{d}{d q} [f (g (q))]$ is $f' (g (q)) g' (q)$ where $f (q) = e^{q}$ and $g (q) = - 0.2 q$ .

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Step 3.2.1

To apply the Chain Rule, set $u$ as $- 0.2 q$ .

$400 (\frac{d}{d u} [e^{u}] \frac{d}{d q} [- 0.2 q])$

Step 3.2.2

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

$400 (e^{u} \frac{d}{d q} [- 0.2 q])$

Step 3.2.3

Replace all occurrences of $u$ with $- 0.2 q$ .

$400 (e^{- 0.2 q} \frac{d}{d q} [- 0.2 q])$

Step 3.3

Differentiate.

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Step 3.3.1

Since $- 0.2$ is constant with respect to $q$ , the derivative of $- 0.2 q$ with respect to $q$ is $- 0.2 \frac{d}{d q} [q]$ .

$400 e^{- 0.2 q} (- 0.2 \frac{d}{d q} [q])$

Step 3.3.2

Multiply $- 0.2$ by $400$ .

$- 80 e^{- 0.2 q} \frac{d}{d q} [q]$

Step 3.3.3

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

$- 80 e^{- 0.2 q} \cdot 1$

Step 3.3.4

Multiply $- 80$ by $1$ .

$- 80 e^{- 0.2 q}$

Step 4

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

$p' = - 80 e^{- 0.2 q}$

Step 5

Replace $p'$ with $\frac{d p}{d q}$ .

$\frac{d p}{d q} = - 80 e^{- 0.2 q}$