Algebra Examples

${(1 - 2 x)}^{n}$

Step 1

Find the first derivative.

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Step 1.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{n}$ and $g (x) = 1 - 2 x$ .

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Step 1.1.1

To apply the Chain Rule, set $u$ as $1 - 2 x$ .

$f' (x) = \frac{d}{d u} (u^{n}) \frac{d}{d x} (1 - 2 x)$

Step 1.1.2

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

$f' (x) = n u^{n - 1} \frac{d}{d x} (1 - 2 x)$

Step 1.1.3

Replace all occurrences of $u$ with $1 - 2 x$ .

$f' (x) = n {(1 - 2 x)}^{n - 1} \frac{d}{d x} (1 - 2 x)$

Step 1.2

Differentiate.

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Step 1.2.1

By the Sum Rule, the derivative of $1 - 2 x$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [- 2 x]$ .

$f' (x) = n {(1 - 2 x)}^{n - 1} (\frac{d}{d x} (1) + \frac{d}{d x} (- 2 x))$

Step 1.2.2

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

$f' (x) = n {(1 - 2 x)}^{n - 1} (0 + \frac{d}{d x} (- 2 x))$

Step 1.2.3

Add $0$ and $\frac{d}{d x} [- 2 x]$ .

$f' (x) = n {(1 - 2 x)}^{n - 1} \frac{d}{d x} (- 2 x)$

Step 1.2.4

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

$f' (x) = n {(1 - 2 x)}^{n - 1} (- 2 \frac{d}{d x} x)$

Step 1.2.5

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

$f' (x) = n {(1 - 2 x)}^{n - 1} (- 2 \cdot 1)$

Step 1.2.6

Simplify the expression.

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Step 1.2.6.1

Multiply $- 2$ by $1$ .

$f' (x) = n {(1 - 2 x)}^{n - 1} \cdot - 2$

Step 1.2.6.2

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

$f' (x) = - 2 n {(1 - 2 x)}^{n - 1}$

Step 2

Find the second derivative.

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Step 2.1

Since $- 2 n$ is constant with respect to $x$ , the derivative of $- 2 n {(1 - 2 x)}^{n - 1}$ with respect to $x$ is $- 2 n \frac{d}{d x} [{(1 - 2 x)}^{n - 1}]$ .

$f'' (x) = - 2 n \frac{d}{d x} {(1 - 2 x)}^{n - 1}$

Step 2.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{n - 1}$ and $g (x) = 1 - 2 x$ .

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Step 2.2.1

To apply the Chain Rule, set $u$ as $1 - 2 x$ .

$f'' (x) = - 2 n (\frac{d}{d u} (u^{n - 1}) \frac{d}{d x} (1 - 2 x))$

Step 2.2.2

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

$f'' (x) = - 2 n ((n - 1) u^{n - 1 - 1} \frac{d}{d x} (1 - 2 x))$

Step 2.2.3

Replace all occurrences of $u$ with $1 - 2 x$ .

$f'' (x) = - 2 n ((n - 1) {(1 - 2 x)}^{n - 1 - 1} \frac{d}{d x} (1 - 2 x))$

Step 2.3

Differentiate.

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Step 2.3.1

Subtract $1$ from $- 1$ .

$f'' (x) = - 2 n ((n - 1) {(1 - 2 x)}^{n - 2} \frac{d}{d x} (1 - 2 x))$

Step 2.3.2

By the Sum Rule, the derivative of $1 - 2 x$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [- 2 x]$ .

$f'' (x) = - 2 n ((n - 1) {(1 - 2 x)}^{n - 2} (\frac{d}{d x} (1) + \frac{d}{d x} (- 2 x)))$

Step 2.3.3

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

$f'' (x) = - 2 n ((n - 1) {(1 - 2 x)}^{n - 2} (0 + \frac{d}{d x} (- 2 x)))$

Step 2.3.4

Add $0$ and $\frac{d}{d x} [- 2 x]$ .

$f'' (x) = - 2 n ((n - 1) {(1 - 2 x)}^{n - 2} \frac{d}{d x} (- 2 x))$

Step 2.3.5

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

$f'' (x) = - 2 n ((n - 1) {(1 - 2 x)}^{n - 2} (- 2 \frac{d}{d x} x))$

Step 2.3.6

Multiply $- 2$ by $- 2$ .

$f'' (x) = 4 n ((n - 1) {(1 - 2 x)}^{n - 2} (\frac{d}{d x} (x)))$

Step 2.3.7

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

$f'' (x) = 4 n ((n - 1) {(1 - 2 x)}^{n - 2} \cdot 1)$

Step 2.3.8

Simplify the expression.

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Step 2.3.8.1

Multiply $n - 1$ by $1$ .

$f'' (x) = 4 n ((n - 1) {(1 - 2 x)}^{n - 2})$

Step 2.3.8.2

Reorder the factors of $4 n ((n - 1) {(1 - 2 x)}^{n - 2})$ .

$f'' (x) = 4 n {(1 - 2 x)}^{n - 2} (n - 1)$

Step 3

Find the third derivative.

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Step 3.1

Since $4 n (n - 1)$ is constant with respect to $x$ , the derivative of $4 n {(1 - 2 x)}^{n - 2} (n - 1)$ with respect to $x$ is $4 n (n - 1) \frac{d}{d x} [{(1 - 2 x)}^{n - 2}]$ .

$f''' (x) = 4 n (n - 1) \frac{d}{d x} ({(1 - 2 x)}^{n - 2})$

Step 3.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{n - 2}$ and $g (x) = 1 - 2 x$ .

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Step 3.2.1

To apply the Chain Rule, set $u$ as $1 - 2 x$ .

$f''' (x) = 4 n (n - 1) (\frac{d}{d u} (u^{n - 2}) \frac{d}{d x} (1 - 2 x))$

Step 3.2.2

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

$f''' (x) = 4 n (n - 1) ((n - 2) u^{n - 2 - 1} \frac{d}{d x} (1 - 2 x))$

Step 3.2.3

Replace all occurrences of $u$ with $1 - 2 x$ .

$f''' (x) = 4 n (n - 1) ((n - 2) {(1 - 2 x)}^{n - 2 - 1} \frac{d}{d x} (1 - 2 x))$

Step 3.3

Differentiate.

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Step 3.3.1

Subtract $1$ from $- 2$ .

$f''' (x) = 4 n (n - 1) ((n - 2) {(1 - 2 x)}^{n - 3} \frac{d}{d x} (1 - 2 x))$

Step 3.3.2

By the Sum Rule, the derivative of $1 - 2 x$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [- 2 x]$ .

$f''' (x) = 4 n (n - 1) ((n - 2) {(1 - 2 x)}^{n - 3} (\frac{d}{d x} (1) + \frac{d}{d x} (- 2 x)))$

Step 3.3.3

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

$f''' (x) = 4 n (n - 1) ((n - 2) {(1 - 2 x)}^{n - 3} (0 + \frac{d}{d x} (- 2 x)))$

Step 3.3.4

Add $0$ and $\frac{d}{d x} [- 2 x]$ .

$f''' (x) = 4 n (n - 1) ((n - 2) {(1 - 2 x)}^{n - 3} \frac{d}{d x} (- 2 x))$

Step 3.3.5

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

$f''' (x) = 4 n (n - 1) ((n - 2) {(1 - 2 x)}^{n - 3} (- 2 \frac{d}{d x} x))$

Step 3.3.6

Multiply $- 2$ by $4$ .

$f''' (x) = - 8 n (n - 1) ((n - 2) {(1 - 2 x)}^{n - 3} (\frac{d}{d x} (x)))$

Step 3.3.7

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

$f''' (x) = - 8 n (n - 1) ((n - 2) {(1 - 2 x)}^{n - 3} \cdot 1)$

Step 3.3.8

Multiply $n - 2$ by $1$ .

$f''' (x) = - 8 n (n - 1) ((n - 2) {(1 - 2 x)}^{n - 3})$

Step 3.4

Simplify.

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Step 3.4.1

Apply the distributive property.

$f''' (x) = (- 8 n \cdot n - 8 n \cdot - 1) ((n - 2) {(1 - 2 x)}^{n - 3})$

Step 3.4.2

Combine terms.

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Step 3.4.2.1

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

$f''' (x) = (- 8 (n \cdot n) - 8 n \cdot - 1) ((n - 2) {(1 - 2 x)}^{n - 3})$

Step 3.4.2.2

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

$f''' (x) = (- 8 (n \cdot n) - 8 n \cdot - 1) ((n - 2) {(1 - 2 x)}^{n - 3})$

Step 3.4.2.3

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

$f''' (x) = (- 8 n^{1 + 1} - 8 n \cdot - 1) ((n - 2) {(1 - 2 x)}^{n - 3})$

Step 3.4.2.4

Add $1$ and $1$ .

$f''' (x) = (- 8 n^{2} - 8 n \cdot - 1) ((n - 2) {(1 - 2 x)}^{n - 3})$

Step 3.4.2.5

Multiply $- 1$ by $- 8$ .

$f''' (x) = (- 8 n^{2} + 8 n) (n - 2) {(1 - 2 x)}^{n - 3}$

Step 3.4.3

Reorder the factors of $(- 8 n^{2} + 8 n) (n - 2) {(1 - 2 x)}^{n - 3}$ .

$f''' (x) = {(1 - 2 x)}^{n - 3} (- 8 n^{2} + 8 n) (n - 2)$

Step 4

Find the fourth derivative.

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Step 4.1

Since $(- 8 n^{2} + 8 n) (n - 2)$ is constant with respect to $x$ , the derivative of ${(1 - 2 x)}^{n - 3} (- 8 n^{2} + 8 n) (n - 2)$ with respect to $x$ is $(- 8 n^{2} + 8 n) (n - 2) \frac{d}{d x} [{(1 - 2 x)}^{n - 3}]$ .

$f^{4} (x) = (- 8 n^{2} + 8 n) (n - 2) \frac{d}{d x} ({(1 - 2 x)}^{n - 3})$

Step 4.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{n - 3}$ and $g (x) = 1 - 2 x$ .

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Step 4.2.1

To apply the Chain Rule, set $u$ as $1 - 2 x$ .

$f^{4} (x) = (- 8 n^{2} + 8 n) (n - 2) (\frac{d}{d u} (u^{n - 3}) \frac{d}{d x} (1 - 2 x))$

Step 4.2.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = n - 3$ .

$f^{4} (x) = (- 8 n^{2} + 8 n) (n - 2) ((n - 3) u^{n - 3 - 1} \frac{d}{d x} (1 - 2 x))$

Step 4.2.3

Replace all occurrences of $u$ with $1 - 2 x$ .

$f^{4} (x) = (- 8 n^{2} + 8 n) (n - 2) ((n - 3) {(1 - 2 x)}^{n - 3 - 1} \frac{d}{d x} (1 - 2 x))$

Step 4.3

Differentiate.

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Step 4.3.1

Subtract $1$ from $- 3$ .

$f^{4} (x) = (- 8 n^{2} + 8 n) (n - 2) ((n - 3) {(1 - 2 x)}^{n - 4} \frac{d}{d x} (1 - 2 x))$

Step 4.3.2

By the Sum Rule, the derivative of $1 - 2 x$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [- 2 x]$ .

$f^{4} (x) = (- 8 n^{2} + 8 n) (n - 2) ((n - 3) {(1 - 2 x)}^{n - 4} (\frac{d}{d x} (1) + \frac{d}{d x} (- 2 x)))$

Step 4.3.3

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

$f^{4} (x) = (- 8 n^{2} + 8 n) (n - 2) ((n - 3) {(1 - 2 x)}^{n - 4} (0 + \frac{d}{d x} (- 2 x)))$

Step 4.3.4

Add $0$ and $\frac{d}{d x} [- 2 x]$ .

$f^{4} (x) = (- 8 n^{2} + 8 n) (n - 2) ((n - 3) {(1 - 2 x)}^{n - 4} \frac{d}{d x} (- 2 x))$

Step 4.3.5

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

$f^{4} (x) = (- 8 n^{2} + 8 n) (n - 2) ((n - 3) {(1 - 2 x)}^{n - 4} (- 2 \frac{d}{d x} x))$

Step 4.3.6

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

$f^{4} (x) = (- 8 n^{2} + 8 n) (n - 2) ((n - 3) {(1 - 2 x)}^{n - 4} (- 2 \cdot 1))$

Step 4.3.7

Simplify the expression.

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Step 4.3.7.1

Multiply $- 2$ by $1$ .

$f^{4} (x) = (- 8 n^{2} + 8 n) (n - 2) ((n - 3) {(1 - 2 x)}^{n - 4} \cdot - 2)$

Step 4.3.7.2

Move $- 2$ to the left of $(n - 3) {(1 - 2 x)}^{n - 4}$ .

$f^{4} (x) = (- 8 n^{2} + 8 n) (n - 2) (- 2 \cdot ((n - 3) {(1 - 2 x)}^{n - 4}))$

Step 4.4

Simplify.

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Step 4.4.1

Apply the distributive property.

$f^{4} (x) = (- 8 n^{2} + 8 n) (n - 2) ((- 2 n - 2 \cdot - 3) {(1 - 2 x)}^{n - 4})$

Step 4.4.2

Multiply $- 2$ by $- 3$ .

$f^{4} (x) = (- 8 n^{2} + 8 n) (n - 2) ((- 2 n + 6) {(1 - 2 x)}^{n - 4})$

Step 4.4.3

Reorder the factors of $(- 8 n^{2} + 8 n) (n - 2) ((- 2 n + 6) {(1 - 2 x)}^{n - 4})$ .

$f^{4} (x) = {(1 - 2 x)}^{n - 4} (- 2 n + 6) (- 8 n^{2} + 8 n) (n - 2)$