Calculus Examples

$p = 4 q e^{q^{9}}$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d q} (p) = \frac{d}{d q} (4 q e^{q^{9}})$

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 $4$ is constant with respect to $q$ , the derivative of $4 q e^{q^{9}}$ with respect to $q$ is $4 \frac{d}{d q} [q e^{q^{9}}]$ .

$4 \frac{d}{d q} [q e^{q^{9}}]$

Step 3.2

Differentiate using the Product Rule which states that $\frac{d}{d q} [f (q) g (q)]$ is $f (q) \frac{d}{d q} [g (q)] + g (q) \frac{d}{d q} [f (q)]$ where $f (q) = q$ and $g (q) = e^{q^{9}}$ .

$4 (q \frac{d}{d q} [e^{q^{9}}] + e^{q^{9}} \frac{d}{d q} [q])$

Step 3.3

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 $g (q) = q^{9}$ .

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Step 3.3.1

To apply the Chain Rule, set $u$ as $q^{9}$ .

$4 (q (\frac{d}{d u} [e^{u}] \frac{d}{d q} [q^{9}]) + e^{q^{9}} \frac{d}{d q} [q])$

Step 3.3.2

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

$4 (q (e^{u} \frac{d}{d q} [q^{9}]) + e^{q^{9}} \frac{d}{d q} [q])$

Step 3.3.3

Replace all occurrences of $u$ with $q^{9}$ .

$4 (q (e^{q^{9}} \frac{d}{d q} [q^{9}]) + e^{q^{9}} \frac{d}{d q} [q])$

Step 3.4

Differentiate using the Power Rule which states that $\frac{d}{d q} [q^{n}]$ is $n q^{n - 1}$ where $n = 9$ .

$4 (q (e^{q^{9}} (9 q^{8})) + e^{q^{9}} \frac{d}{d q} [q])$

Step 3.5

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

$4 (q^{8} q^{1} (e^{q^{9}} \cdot (9)) + e^{q^{9}} \frac{d}{d q} [q])$

Step 3.6

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

$4 (q^{8 + 1} (e^{q^{9}} \cdot (9)) + e^{q^{9}} \frac{d}{d q} [q])$

Step 3.7

Simplify the expression.

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Step 3.7.1

Add $8$ and $1$ .

$4 (q^{9} (e^{q^{9}} \cdot (9)) + e^{q^{9}} \frac{d}{d q} [q])$

Step 3.7.2

Move $9$ to the left of $e^{q^{9}}$ .

$4 (q^{9} (9 \cdot e^{q^{9}}) + e^{q^{9}} \frac{d}{d q} [q])$

Step 3.8

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

$4 (q^{9} (9 e^{q^{9}}) + e^{q^{9}} \cdot 1)$

Step 3.9

Multiply $e^{q^{9}}$ by $1$ .

$4 (q^{9} (9 e^{q^{9}}) + e^{q^{9}})$

Step 3.10

Simplify.

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Step 3.10.1

Apply the distributive property.

$4 (q^{9} (9 e^{q^{9}})) + 4 e^{q^{9}}$

Step 3.10.2

Multiply $9$ by $4$ .

$36 q^{9} e^{q^{9}} + 4 e^{q^{9}}$

Step 3.10.3

Reorder terms.

$36 e^{q^{9}} q^{9} + 4 e^{q^{9}}$

Step 3.10.4

Reorder factors in $36 e^{q^{9}} q^{9} + 4 e^{q^{9}}$ .

$36 q^{9} e^{q^{9}} + 4 e^{q^{9}}$

Step 4

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

$p' = 36 q^{9} e^{q^{9}} + 4 e^{q^{9}}$

Step 5

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

$\frac{d p}{d q} = 36 q^{9} e^{q^{9}} + 4 e^{q^{9}}$