Find the Fourth Derivative (x-3)^7

${(x - 3)}^{7}$

Step 1

Find the first derivative.

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{7}$ and $g (x) = x - 3$ .

To apply the Chain Rule, set $u$ as $x - 3$ .

$f' (x) = \frac{d}{d u} (u^{7}) \frac{d}{d x} (x - 3)$

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = 7$ .

$f' (x) = 7 u^{6} \frac{d}{d x} (x - 3)$

Replace all occurrences of $u$ with $x - 3$ .

$f' (x) = 7 {(x - 3)}^{6} \frac{d}{d x} (x - 3)$

Differentiate.

By the Sum Rule, the derivative of $x - 3$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 3]$ .

$f' (x) = 7 {(x - 3)}^{6} (\frac{d}{d x} (x) + \frac{d}{d x} (- 3))$

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

$f' (x) = 7 {(x - 3)}^{6} (1 + \frac{d}{d x} (- 3))$

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3$ with respect to $x$ is $0$ .

$f' (x) = 7 {(x - 3)}^{6} (1 + 0)$

Simplify the expression.

Add $1$ and $0$ .

$f' (x) = 7 {(x - 3)}^{6} \cdot 1$

Multiply $7$ by $1$ .

$f' (x) = 7 {(x - 3)}^{6}$

Step 2

Find the second derivative.

Since $7$ is constant with respect to $x$ , the derivative of $7 {(x - 3)}^{6}$ with respect to $x$ is $7 \frac{d}{d x} [{(x - 3)}^{6}]$ .

$f'' (x) = 7 \frac{d}{d x} ({(x - 3)}^{6})$

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{6}$ and $g (x) = x - 3$ .

To apply the Chain Rule, set $u$ as $x - 3$ .

$f'' (x) = 7 (\frac{d}{d u} (u^{6}) \frac{d}{d x} (x - 3))$

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = 6$ .

$f'' (x) = 7 (6 u^{5} \frac{d}{d x} (x - 3))$

Replace all occurrences of $u$ with $x - 3$ .

$f'' (x) = 7 (6 {(x - 3)}^{5} \frac{d}{d x} (x - 3))$

Differentiate.

Multiply $6$ by $7$ .

$f'' (x) = 42 ({(x - 3)}^{5} \frac{d}{d x} (x - 3))$

By the Sum Rule, the derivative of $x - 3$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 3]$ .

$f'' (x) = 42 {(x - 3)}^{5} (\frac{d}{d x} (x) + \frac{d}{d x} (- 3))$

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

$f'' (x) = 42 {(x - 3)}^{5} (1 + \frac{d}{d x} (- 3))$

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3$ with respect to $x$ is $0$ .

$f'' (x) = 42 {(x - 3)}^{5} (1 + 0)$

Simplify the expression.

Add $1$ and $0$ .

$f'' (x) = 42 {(x - 3)}^{5} \cdot 1$

Multiply $42$ by $1$ .

$f'' (x) = 42 {(x - 3)}^{5}$

Step 3

Find the third derivative.

Since $42$ is constant with respect to $x$ , the derivative of $42 {(x - 3)}^{5}$ with respect to $x$ is $42 \frac{d}{d x} [{(x - 3)}^{5}]$ .

$f''' (x) = 42 \frac{d}{d x} ({(x - 3)}^{5})$

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{5}$ and $g (x) = x - 3$ .

To apply the Chain Rule, set $u$ as $x - 3$ .

$f''' (x) = 42 (\frac{d}{d u} (u^{5}) \frac{d}{d x} (x - 3))$

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = 5$ .

$f''' (x) = 42 (5 u^{4} \frac{d}{d x} (x - 3))$

Replace all occurrences of $u$ with $x - 3$ .

$f''' (x) = 42 (5 {(x - 3)}^{4} \frac{d}{d x} (x - 3))$

Differentiate.

Multiply $5$ by $42$ .

$f''' (x) = 210 ({(x - 3)}^{4} \frac{d}{d x} (x - 3))$

By the Sum Rule, the derivative of $x - 3$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 3]$ .

$f''' (x) = 210 {(x - 3)}^{4} (\frac{d}{d x} (x) + \frac{d}{d x} (- 3))$

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

$f''' (x) = 210 {(x - 3)}^{4} (1 + \frac{d}{d x} (- 3))$

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3$ with respect to $x$ is $0$ .

$f''' (x) = 210 {(x - 3)}^{4} (1 + 0)$

Simplify the expression.

Add $1$ and $0$ .

$f''' (x) = 210 {(x - 3)}^{4} \cdot 1$

Multiply $210$ by $1$ .

$f''' (x) = 210 {(x - 3)}^{4}$

Step 4

Find the fourth derivative.

Since $210$ is constant with respect to $x$ , the derivative of $210 {(x - 3)}^{4}$ with respect to $x$ is $210 \frac{d}{d x} [{(x - 3)}^{4}]$ .

$f^{4} (x) = 210 \frac{d}{d x} ({(x - 3)}^{4})$

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{4}$ and $g (x) = x - 3$ .

To apply the Chain Rule, set $u$ as $x - 3$ .

$f^{4} (x) = 210 (\frac{d}{d u} (u^{4}) \frac{d}{d x} (x - 3))$

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = 4$ .

$f^{4} (x) = 210 (4 u^{3} \frac{d}{d x} (x - 3))$

Replace all occurrences of $u$ with $x - 3$ .

$f^{4} (x) = 210 (4 {(x - 3)}^{3} \frac{d}{d x} (x - 3))$

Differentiate.

Multiply $4$ by $210$ .

$f^{4} (x) = 840 ({(x - 3)}^{3} \frac{d}{d x} (x - 3))$

By the Sum Rule, the derivative of $x - 3$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 3]$ .

$f^{4} (x) = 840 {(x - 3)}^{3} (\frac{d}{d x} (x) + \frac{d}{d x} (- 3))$

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

$f^{4} (x) = 840 {(x - 3)}^{3} (1 + \frac{d}{d x} (- 3))$

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3$ with respect to $x$ is $0$ .

$f^{4} (x) = 840 {(x - 3)}^{3} (1 + 0)$

Simplify the expression.

Add $1$ and $0$ .

$f^{4} (x) = 840 {(x - 3)}^{3} \cdot 1$

Multiply $840$ by $1$ .

$f^{4} (x) = 840 {(x - 3)}^{3}$

Find the Fourth Derivative (x-3)^7

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