Calculus Examples

$83 - 16 \ln (t + 1)$

Step 1

Differentiate.

Tap for more steps...

Step 1.1

By the Sum Rule, the derivative of $83 - 16 \ln (t + 1)$ with respect to $t$ is $\frac{d}{d t} [83] + \frac{d}{d t} [- 16 \ln (t + 1)]$ .

$\frac{d}{d t} [83] + \frac{d}{d t} [- 16 \ln (t + 1)]$

Step 1.2

Since $83$ is constant with respect to $t$ , the derivative of $83$ with respect to $t$ is $0$ .

$0 + \frac{d}{d t} [- 16 \ln (t + 1)]$

Step 2

Evaluate $\frac{d}{d t} [- 16 \ln (t + 1)]$ .

Tap for more steps...

Step 2.1

Since $- 16$ is constant with respect to $t$ , the derivative of $- 16 \ln (t + 1)$ with respect to $t$ is $- 16 \frac{d}{d t} [\ln (t + 1)]$ .

$0 - 16 \frac{d}{d t} [\ln (t + 1)]$

Step 2.2

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 $g (t) = t + 1$ .

Tap for more steps...

Step 2.2.1

To apply the Chain Rule, set $u$ as $t + 1$ .

$0 - 16 (\frac{d}{d u} [\ln (u)] \frac{d}{d t} [t + 1])$

Step 2.2.2

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

$0 - 16 (\frac{1}{u} \frac{d}{d t} [t + 1])$

Step 2.2.3

Replace all occurrences of $u$ with $t + 1$ .

$0 - 16 (\frac{1}{t + 1} \frac{d}{d t} [t + 1])$

Step 2.3

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

$0 - 16 (\frac{1}{t + 1} (\frac{d}{d t} [t] + \frac{d}{d t} [1]))$

Step 2.4

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

$0 - 16 (\frac{1}{t + 1} (1 + \frac{d}{d t} [1]))$

Step 2.5

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

$0 - 16 (\frac{1}{t + 1} (1 + 0))$

Step 2.6

Add $1$ and $0$ .

$0 - 16 (\frac{1}{t + 1} \cdot 1)$

Step 2.7

Multiply $\frac{1}{t + 1}$ by $1$ .

$0 - 16 \frac{1}{t + 1}$

Step 2.8

Combine $- 16$ and $\frac{1}{t + 1}$ .

$0 + \frac{- 16}{t + 1}$

Step 2.9

Move the negative in front of the fraction.

$0 - \frac{16}{t + 1}$

Step 3

Subtract $\frac{16}{t + 1}$ from $0$ .

$- \frac{16}{t + 1}$