Calculus Examples

$x^{4} \ln (x) - \frac{1}{2} x^{2}$

Step 1

By the Sum Rule, the derivative of $x^{4} \ln (x) - \frac{1}{2} x^{2}$ with respect to $x$ is $\frac{d}{d x} [x^{4} \ln (x)] + \frac{d}{d x} [- \frac{1}{2} x^{2}]$ .

$\frac{d}{d x} [x^{4} \ln (x)] + \frac{d}{d x} [- \frac{1}{2} x^{2}]$

Step 2

Evaluate $\frac{d}{d x} [x^{4} \ln (x)]$ .

Tap for more steps...

Step 2.1

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x^{4}$ and $g (x) = \ln (x)$ .

$x^{4} \frac{d}{d x} [\ln (x)] + \ln (x) \frac{d}{d x} [x^{4}] + \frac{d}{d x} [- \frac{1}{2} x^{2}]$

Step 2.2

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

$x^{4} \frac{1}{x} + \ln (x) \frac{d}{d x} [x^{4}] + \frac{d}{d x} [- \frac{1}{2} x^{2}]$

Step 2.3

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

$x^{4} \frac{1}{x} + \ln (x) (4 x^{3}) + \frac{d}{d x} [- \frac{1}{2} x^{2}]$

Step 2.4

Combine $x^{4}$ and $\frac{1}{x}$ .

$\frac{x^{4}}{x} + \ln (x) (4 x^{3}) + \frac{d}{d x} [- \frac{1}{2} x^{2}]$

Step 2.5

Cancel the common factor of $x^{4}$ and $x$ .

Tap for more steps...

Step 2.5.1

Factor $x$ out of $x^{4}$ .

$\frac{x \cdot x^{3}}{x} + \ln (x) (4 x^{3}) + \frac{d}{d x} [- \frac{1}{2} x^{2}]$

Step 2.5.2

Cancel the common factors.

Tap for more steps...

Step 2.5.2.1

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

$\frac{x \cdot x^{3}}{x^{1}} + \ln (x) (4 x^{3}) + \frac{d}{d x} [- \frac{1}{2} x^{2}]$

Step 2.5.2.2

Factor $x$ out of $x^{1}$ .

$\frac{x \cdot x^{3}}{x \cdot 1} + \ln (x) (4 x^{3}) + \frac{d}{d x} [- \frac{1}{2} x^{2}]$

Step 2.5.2.3

Cancel the common factor.

$\frac{x \cdot x^{3}}{x \cdot 1} + \ln (x) (4 x^{3}) + \frac{d}{d x} [- \frac{1}{2} x^{2}]$

Step 2.5.2.4

Rewrite the expression.

$\frac{x^{3}}{1} + \ln (x) (4 x^{3}) + \frac{d}{d x} [- \frac{1}{2} x^{2}]$

Step 2.5.2.5

Divide $x^{3}$ by $1$ .

$x^{3} + \ln (x) (4 x^{3}) + \frac{d}{d x} [- \frac{1}{2} x^{2}]$

Step 3

Evaluate $\frac{d}{d x} [- \frac{1}{2} x^{2}]$ .

Tap for more steps...

Step 3.1

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

$x^{3} + \ln (x) (4 x^{3}) - \frac{1}{2} \frac{d}{d x} [x^{2}]$

Step 3.2

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

$x^{3} + \ln (x) (4 x^{3}) - \frac{1}{2} (2 x)$

Step 3.3

Multiply $2$ by $- 1$ .

$x^{3} + \ln (x) (4 x^{3}) - 2 (\frac{1}{2}) x$

Step 3.4

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

$x^{3} + \ln (x) (4 x^{3}) + \frac{- 2}{2} x$

Step 3.5

Combine $\frac{- 2}{2}$ and $x$ .

$x^{3} + \ln (x) (4 x^{3}) + \frac{- 2 x}{2}$

Step 3.6

Cancel the common factor of $- 2$ and $2$ .

Tap for more steps...

Step 3.6.1

Factor $2$ out of $- 2 x$ .

$x^{3} + \ln (x) (4 x^{3}) + \frac{2 (- x)}{2}$

Step 3.6.2

Cancel the common factors.

Tap for more steps...

Step 3.6.2.1

Factor $2$ out of $2$ .

$x^{3} + \ln (x) (4 x^{3}) + \frac{2 (- x)}{2 (1)}$

Step 3.6.2.2

Cancel the common factor.

$x^{3} + \ln (x) (4 x^{3}) + \frac{2 (- x)}{2 \cdot 1}$

Step 3.6.2.3

Rewrite the expression.

$x^{3} + \ln (x) (4 x^{3}) + \frac{- x}{1}$

Step 3.6.2.4

Divide $- x$ by $1$ .

$x^{3} + \ln (x) (4 x^{3}) - x$

Step 4

Reorder terms.

$x^{3} + 4 x^{3} \ln (x) - x$