Calculus Examples

$y = {(2 x - 7)}^{3}$

Step 1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{3}$ and $g (x) = 2 x - 7$ .

Tap for more steps...

Step 1.1

To apply the Chain Rule, set $u$ as $2 x - 7$ .

$\frac{d}{d u} [u^{3}] \frac{d}{d x} [2 x - 7]$

Step 1.2

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

$3 u^{2} \frac{d}{d x} [2 x - 7]$

Step 1.3

Replace all occurrences of $u$ with $2 x - 7$ .

$3 {(2 x - 7)}^{2} \frac{d}{d x} [2 x - 7]$

Step 2

By the Sum Rule, the derivative of $2 x - 7$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [- 7]$ .

$3 {(2 x - 7)}^{2} (\frac{d}{d x} [2 x] + \frac{d}{d x} [- 7])$

Step 3

Evaluate $\frac{d}{d x} [2 x]$ .

Tap for more steps...

Step 3.1

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

$3 {(2 x - 7)}^{2} (2 \frac{d}{d x} [x] + \frac{d}{d x} [- 7])$

Step 3.2

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

$3 {(2 x - 7)}^{2} (2 \cdot 1 + \frac{d}{d x} [- 7])$

Step 3.3

Multiply $2$ by $1$ .

$3 {(2 x - 7)}^{2} (2 + \frac{d}{d x} [- 7])$

Step 4

Differentiate using the Constant Rule.

Tap for more steps...

Step 4.1

Since $- 7$ is constant with respect to $x$ , the derivative of $- 7$ with respect to $x$ is $0$ .

$3 {(2 x - 7)}^{2} (2 + 0)$

Step 4.2

Simplify the expression.

Tap for more steps...

Step 4.2.1

Add $2$ and $0$ .

$3 {(2 x - 7)}^{2} \cdot 2$

Step 4.2.2

Multiply $2$ by $3$ .

$6 {(2 x - 7)}^{2}$