Calculus Examples

$y = \ln ({(1 - t)}^{- 2})$

Step 1

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 $g (t) = {(1 - t)}^{- 2}$ .

Tap for more steps...

Step 1.1

To apply the Chain Rule, set $u_{1}$ as ${(1 - t)}^{- 2}$ .

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

Step 1.2

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

$\frac{1}{u_{1}} \frac{d}{d t} [{(1 - t)}^{- 2}]$

Step 1.3

Replace all occurrences of $u_{1}$ with ${(1 - t)}^{- 2}$ .

$\frac{1}{{(1 - t)}^{- 2}} \frac{d}{d t} [{(1 - t)}^{- 2}]$

Step 2

Move ${(1 - t)}^{- 2}$ to the numerator using the negative exponent rule $\frac{1}{b^{- n}} = b^{n}$ .

${(1 - t)}^{2} \frac{d}{d t} [{(1 - t)}^{- 2}]$

Step 3

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = t^{- 2}$ and $g (t) = 1 - t$ .

Tap for more steps...

Step 3.1

To apply the Chain Rule, set $u_{2}$ as $1 - t$ .

${(1 - t)}^{2} (\frac{d}{d u_{2}} [{u_{2}}^{- 2}] \frac{d}{d t} [1 - t])$

Step 3.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{2}} [{u_{2}}^{n}]$ is $n {u_{2}}^{n - 1}$ where $n = - 2$ .

${(1 - t)}^{2} (- 2 {u_{2}}^{- 3} \frac{d}{d t} [1 - t])$

Step 3.3

Replace all occurrences of $u_{2}$ with $1 - t$ .

${(1 - t)}^{2} (- 2 {(1 - t)}^{- 3} \frac{d}{d t} [1 - t])$

Step 4

Multiply ${(1 - t)}^{2}$ by ${(1 - t)}^{- 3}$ by adding the exponents.

Tap for more steps...

Step 4.1

Move ${(1 - t)}^{- 3}$ .

${(1 - t)}^{- 3} {(1 - t)}^{2} (- 2 \frac{d}{d t} [1 - t])$

Step 4.2

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

${(1 - t)}^{- 3 + 2} (- 2 \frac{d}{d t} [1 - t])$

Step 4.3

Add $- 3$ and $2$ .

${(1 - t)}^{- 1} (- 2 \frac{d}{d t} [1 - t])$

Step 5

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

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

Step 6

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

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

Step 7

Add $0$ and $\frac{d}{d t} [- t]$ .

${(1 - t)}^{- 1} (- 2 \frac{d}{d t} [- t])$

Step 8

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

${(1 - t)}^{- 1} (- 2 (- \frac{d}{d t} [t]))$

Step 9

Simplify the expression.

Tap for more steps...

Step 9.1

Multiply $- 1$ by $- 2$ .

${(1 - t)}^{- 1} (2 \frac{d}{d t} [t])$

Step 9.2

Move $2$ to the left of ${(1 - t)}^{- 1}$ .

$2 \cdot {(1 - t)}^{- 1} \frac{d}{d t} [t]$

Step 10

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

$2 {(1 - t)}^{- 1} \cdot 1$

Step 11

Multiply $2$ by $1$ .

$2 {(1 - t)}^{- 1}$

Step 12

Simplify.

Tap for more steps...

Step 12.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$2 \frac{1}{1 - t}$

Step 12.2

Combine $2$ and $\frac{1}{1 - t}$ .

$\frac{2}{1 - t}$