Calculus Examples

$f (t) = e^{2 t} \ln (t + 1)$

Step 1

Differentiate using the Product Rule which states that $\frac{d}{d t} [f (t) g (t)]$ is $f (t) \frac{d}{d t} [g (t)] + g (t) \frac{d}{d t} [f (t)]$ where $f (t) = e^{2 t}$ and $g (t) = \ln (t + 1)$ .

$e^{2 t} \frac{d}{d t} [\ln (t + 1)] + \ln (t + 1) \frac{d}{d t} [e^{2 t}]$

Step 2

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = \ln (t)$ and $g (t) = t + 1$ .

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Step 2.1

To apply the Chain Rule, set $u_{1}$ as $t + 1$ .

$e^{2 t} (\frac{d}{d u_{1}} [\ln (u_{1})] \frac{d}{d t} [t + 1]) + \ln (t + 1) \frac{d}{d t} [e^{2 t}]$

Step 2.2

The derivative of $\ln (u_{1})$ with respect to $u_{1}$ is $\frac{1}{u_{1}}$ .

$e^{2 t} (\frac{1}{u_{1}} \frac{d}{d t} [t + 1]) + \ln (t + 1) \frac{d}{d t} [e^{2 t}]$

Step 2.3

Replace all occurrences of $u_{1}$ with $t + 1$ .

$e^{2 t} (\frac{1}{t + 1} \frac{d}{d t} [t + 1]) + \ln (t + 1) \frac{d}{d t} [e^{2 t}]$

Step 3

Differentiate.

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Step 3.1

Combine $\frac{1}{t + 1}$ and $e^{2 t}$ .

$\frac{e^{2 t}}{t + 1} \frac{d}{d t} [t + 1] + \ln (t + 1) \frac{d}{d t} [e^{2 t}]$

Step 3.2

By the Sum Rule, the derivative of $t + 1$ with respect to $t$ is $\frac{d}{d t} [t] + \frac{d}{d t} [1]$ .

$\frac{e^{2 t}}{t + 1} (\frac{d}{d t} [t] + \frac{d}{d t} [1]) + \ln (t + 1) \frac{d}{d t} [e^{2 t}]$

Step 3.3

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

$\frac{e^{2 t}}{t + 1} (1 + \frac{d}{d t} [1]) + \ln (t + 1) \frac{d}{d t} [e^{2 t}]$

Step 3.4

Since $1$ is constant with respect to $t$ , the derivative of $1$ with respect to $t$ is $0$ .

$\frac{e^{2 t}}{t + 1} (1 + 0) + \ln (t + 1) \frac{d}{d t} [e^{2 t}]$

Step 3.5

Simplify the expression.

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Step 3.5.1

Add $1$ and $0$ .

$\frac{e^{2 t}}{t + 1} \cdot 1 + \ln (t + 1) \frac{d}{d t} [e^{2 t}]$

Step 3.5.2

Multiply $\frac{e^{2 t}}{t + 1}$ by $1$ .

$\frac{e^{2 t}}{t + 1} + \ln (t + 1) \frac{d}{d t} [e^{2 t}]$

Step 4

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = e^{t}$ and $g (t) = 2 t$ .

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Step 4.1

To apply the Chain Rule, set $u_{2}$ as $2 t$ .

$\frac{e^{2 t}}{t + 1} + \ln (t + 1) (\frac{d}{d u_{2}} [e^{u_{2}}] \frac{d}{d t} [2 t])$

Step 4.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{2}} [a^{u_{2}}]$ is $a^{u_{2}} \ln (a)$ where $a$ = $e$ .

$\frac{e^{2 t}}{t + 1} + \ln (t + 1) (e^{u_{2}} \frac{d}{d t} [2 t])$

Step 4.3

Replace all occurrences of $u_{2}$ with $2 t$ .

$\frac{e^{2 t}}{t + 1} + \ln (t + 1) (e^{2 t} \frac{d}{d t} [2 t])$

Step 5

Differentiate.

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Step 5.1

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

$\frac{e^{2 t}}{t + 1} + \ln (t + 1) (e^{2 t} (2 \frac{d}{d t} [t]))$

Step 5.2

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

$\frac{e^{2 t}}{t + 1} + \ln (t + 1) (e^{2 t} (2 \cdot 1))$

Step 5.3

Simplify the expression.

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Step 5.3.1

Multiply $2$ by $1$ .

$\frac{e^{2 t}}{t + 1} + \ln (t + 1) (e^{2 t} \cdot 2)$

Step 5.3.2

Move $2$ to the left of $e^{2 t}$ .

$\frac{e^{2 t}}{t + 1} + \ln (t + 1) (2 \cdot e^{2 t})$

Step 6

To write $\ln (t + 1) (2 e^{2 t})$ as a fraction with a common denominator, multiply by $\frac{t + 1}{t + 1}$ .

$\frac{e^{2 t}}{t + 1} + \frac{\ln (t + 1) (2 e^{2 t}) (t + 1)}{t + 1}$

Step 7

Combine the numerators over the common denominator.

$\frac{e^{2 t} + \ln (t + 1) (2 e^{2 t}) (t + 1)}{t + 1}$

Step 8

Simplify.

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Step 8.1

Simplify the numerator.

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Step 8.1.1

Simplify each term.

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Step 8.1.1.1

Rewrite using the commutative property of multiplication.

$\frac{e^{2 t} + 2 \ln (t + 1) e^{2 t} (t + 1)}{t + 1}$

Step 8.1.1.2

Simplify $2 \ln (t + 1)$ by moving $2$ inside the logarithm.

$\frac{e^{2 t} + \ln ({(t + 1)}^{2}) e^{2 t} (t + 1)}{t + 1}$

Step 8.1.1.3

Apply the distributive property.

$\frac{e^{2 t} + \ln ({(t + 1)}^{2}) e^{2 t} t + \ln ({(t + 1)}^{2}) e^{2 t} \cdot 1}{t + 1}$

Step 8.1.1.4

Multiply $\ln ({(t + 1)}^{2})$ by $1$ .

$\frac{e^{2 t} + \ln ({(t + 1)}^{2}) e^{2 t} t + \ln ({(t + 1)}^{2}) e^{2 t}}{t + 1}$

Step 8.1.2

Reorder factors in $e^{2 t} + \ln ({(t + 1)}^{2}) e^{2 t} t + \ln ({(t + 1)}^{2}) e^{2 t}$ .

$\frac{e^{2 t} + t e^{2 t} \ln ({(t + 1)}^{2}) + e^{2 t} \ln ({(t + 1)}^{2})}{t + 1}$

Step 8.2

Reorder terms.

$\frac{e^{2 t} t \ln ({(t + 1)}^{2}) + e^{2 t} \ln ({(t + 1)}^{2}) + e^{2 t}}{t + 1}$