Calculus Examples

$5 \ln (5 \ln (4 t))$

Step 1

Since $5$ is constant with respect to $t$ , the derivative of $5 \ln (5 \ln (4 t))$ with respect to $t$ is $5 \frac{d}{d t} [\ln (5 \ln (4 t))]$ .

$5 \frac{d}{d t} [\ln (5 \ln (4 t))]$

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 $g (t) = 5 \ln (4 t)$ .

Tap for more steps...

Step 2.1

To apply the Chain Rule, set $u_{1}$ as $5 \ln (4 t)$ .

$5 (\frac{d}{d u_{1}} [\ln (u_{1})] \frac{d}{d t} [5 \ln (4 t)])$

Step 2.2

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

$5 (\frac{1}{u_{1}} \frac{d}{d t} [5 \ln (4 t)])$

Step 2.3

Replace all occurrences of $u_{1}$ with $5 \ln (4 t)$ .

$5 (\frac{1}{5 \ln (4 t)} \frac{d}{d t} [5 \ln (4 t)])$

Step 3

Differentiate using the Constant Multiple Rule.

Tap for more steps...

Step 3.1

Combine $\frac{1}{5 \ln (4 t)}$ and $5$ .

$\frac{5}{5 \ln (4 t)} \frac{d}{d t} [5 \ln (4 t)]$

Step 3.2

Cancel the common factor of $5$ .

Tap for more steps...

Step 3.2.1

Cancel the common factor.

$\frac{5}{5 \ln (4 t)} \frac{d}{d t} [5 \ln (4 t)]$

Step 3.2.2

Rewrite the expression.

$\frac{1}{\ln (4 t)} \frac{d}{d t} [5 \ln (4 t)]$

Step 3.3

Since $5$ is constant with respect to $t$ , the derivative of $5 \ln (4 t)$ with respect to $t$ is $5 \frac{d}{d t} [\ln (4 t)]$ .

$\frac{1}{\ln (4 t)} (5 \frac{d}{d t} [\ln (4 t)])$

Step 3.4

Combine $5$ and $\frac{1}{\ln (4 t)}$ .

$\frac{5}{\ln (4 t)} \frac{d}{d t} [\ln (4 t)]$

Step 4

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = \ln (t)$ and $g (t) = 4 t$ .

Tap for more steps...

Step 4.1

To apply the Chain Rule, set $u_{2}$ as $4 t$ .

$\frac{5}{\ln (4 t)} (\frac{d}{d u_{2}} [\ln (u_{2})] \frac{d}{d t} [4 t])$

Step 4.2

The derivative of $\ln (u_{2})$ with respect to $u_{2}$ is $\frac{1}{u_{2}}$ .

$\frac{5}{\ln (4 t)} (\frac{1}{u_{2}} \frac{d}{d t} [4 t])$

Step 4.3

Replace all occurrences of $u_{2}$ with $4 t$ .

$\frac{5}{\ln (4 t)} (\frac{1}{4 t} \frac{d}{d t} [4 t])$

Step 5

Differentiate.

Tap for more steps...

Step 5.1

Multiply $\frac{1}{4 t}$ by $\frac{5}{\ln (4 t)}$ .

$\frac{5}{4 t \ln (4 t)} \frac{d}{d t} [4 t]$

Step 5.2

Since $4$ is constant with respect to $t$ , the derivative of $4 t$ with respect to $t$ is $4 \frac{d}{d t} [t]$ .

$\frac{5}{4 t \ln (4 t)} (4 \frac{d}{d t} [t])$

Step 5.3

Simplify terms.

Tap for more steps...

Step 5.3.1

Combine $4$ and $\frac{5}{4 t \ln (4 t)}$ .

$\frac{4 \cdot 5}{4 t \ln (4 t)} \frac{d}{d t} [t]$

Step 5.3.2

Multiply $4$ by $5$ .

$\frac{20}{4 t \ln (4 t)} \frac{d}{d t} [t]$

Step 5.3.3

Cancel the common factor of $20$ and $4$ .

Tap for more steps...

Step 5.3.3.1

Factor $4$ out of $20$ .

$\frac{4 \cdot 5}{4 t \ln (4 t)} \frac{d}{d t} [t]$

Step 5.3.3.2

Cancel the common factors.

Tap for more steps...

Step 5.3.3.2.1

Factor $4$ out of $4 t \ln (4 t)$ .

$\frac{4 \cdot 5}{4 (t \ln (4 t))} \frac{d}{d t} [t]$

Step 5.3.3.2.2

Cancel the common factor.

$\frac{4 \cdot 5}{4 (t \ln (4 t))} \frac{d}{d t} [t]$

Step 5.3.3.2.3

Rewrite the expression.

$\frac{5}{t \ln (4 t)} \frac{d}{d t} [t]$

Step 5.4

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

$\frac{5}{t \ln (4 t)} \cdot 1$

Step 5.5

Multiply $\frac{5}{t \ln (4 t)}$ by $1$ .

$\frac{5}{t \ln (4 t)}$