Find the 4th Derivative 6x^5-5x^4+4x^3+8x^2+8

$6 x^{5} - 5 x^{4} + 4 x^{3} + 8 x^{2} + 8$

Step 1

Find the first derivative.

By the Sum Rule, the derivative of $6 x^{5} - 5 x^{4} + 4 x^{3} + 8 x^{2} + 8$ with respect to $x$ is $\frac{d}{d x} [6 x^{5}] + \frac{d}{d x} [- 5 x^{4}] + \frac{d}{d x} [4 x^{3}] + \frac{d}{d x} [8 x^{2}] + \frac{d}{d x} [8]$ .

$f' (x) = \frac{d}{d x} (6 x^{5}) + \frac{d}{d x} (- 5 x^{4}) + \frac{d}{d x} (4 x^{3}) + \frac{d}{d x} (8 x^{2}) + \frac{d}{d x} (8)$

Evaluate $\frac{d}{d x} [6 x^{5}]$ .

Since $6$ is constant with respect to $x$ , the derivative of $6 x^{5}$ with respect to $x$ is $6 \frac{d}{d x} [x^{5}]$ .

$f' (x) = 6 \frac{d}{d x} (x^{5}) + \frac{d}{d x} (- 5 x^{4}) + \frac{d}{d x} (4 x^{3}) + \frac{d}{d x} (8 x^{2}) + \frac{d}{d x} (8)$

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

$f' (x) = 6 (5 x^{4}) + \frac{d}{d x} (- 5 x^{4}) + \frac{d}{d x} (4 x^{3}) + \frac{d}{d x} (8 x^{2}) + \frac{d}{d x} (8)$

Multiply $5$ by $6$ .

$f' (x) = 30 x^{4} + \frac{d}{d x} (- 5 x^{4}) + \frac{d}{d x} (4 x^{3}) + \frac{d}{d x} (8 x^{2}) + \frac{d}{d x} (8)$

Evaluate $\frac{d}{d x} [- 5 x^{4}]$ .

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5 x^{4}$ with respect to $x$ is $- 5 \frac{d}{d x} [x^{4}]$ .

$f' (x) = 30 x^{4} - 5 \frac{d}{d x} x^{4} + \frac{d}{d x} (4 x^{3}) + \frac{d}{d x} (8 x^{2}) + \frac{d}{d x} (8)$

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

$f' (x) = 30 x^{4} - 5 (4 x^{3}) + \frac{d}{d x} (4 x^{3}) + \frac{d}{d x} (8 x^{2}) + \frac{d}{d x} (8)$

Multiply $4$ by $- 5$ .

$f' (x) = 30 x^{4} - 20 x^{3} + \frac{d}{d x} (4 x^{3}) + \frac{d}{d x} (8 x^{2}) + \frac{d}{d x} (8)$

Evaluate $\frac{d}{d x} [4 x^{3}]$ .

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

$f' (x) = 30 x^{4} - 20 x^{3} + 4 \frac{d}{d x} (x^{3}) + \frac{d}{d x} (8 x^{2}) + \frac{d}{d x} (8)$

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

$f' (x) = 30 x^{4} - 20 x^{3} + 4 (3 x^{2}) + \frac{d}{d x} (8 x^{2}) + \frac{d}{d x} (8)$

Multiply $3$ by $4$ .

$f' (x) = 30 x^{4} - 20 x^{3} + 12 x^{2} + \frac{d}{d x} (8 x^{2}) + \frac{d}{d x} (8)$

Evaluate $\frac{d}{d x} [8 x^{2}]$ .

Since $8$ is constant with respect to $x$ , the derivative of $8 x^{2}$ with respect to $x$ is $8 \frac{d}{d x} [x^{2}]$ .

$f' (x) = 30 x^{4} - 20 x^{3} + 12 x^{2} + 8 \frac{d}{d x} (x^{2}) + \frac{d}{d x} (8)$

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

$f' (x) = 30 x^{4} - 20 x^{3} + 12 x^{2} + 8 (2 x) + \frac{d}{d x} (8)$

Multiply $2$ by $8$ .

$f' (x) = 30 x^{4} - 20 x^{3} + 12 x^{2} + 16 x + \frac{d}{d x} (8)$

Differentiate using the Constant Rule.

Since $8$ is constant with respect to $x$ , the derivative of $8$ with respect to $x$ is $0$ .

$f' (x) = 30 x^{4} - 20 x^{3} + 12 x^{2} + 16 x + 0$

Add $30 x^{4} - 20 x^{3} + 12 x^{2} + 16 x$ and $0$ .

$f' (x) = 30 x^{4} - 20 x^{3} + 12 x^{2} + 16 x$

Step 2

Find the second derivative.

By the Sum Rule, the derivative of $30 x^{4} - 20 x^{3} + 12 x^{2} + 16 x$ with respect to $x$ is $\frac{d}{d x} [30 x^{4}] + \frac{d}{d x} [- 20 x^{3}] + \frac{d}{d x} [12 x^{2}] + \frac{d}{d x} [16 x]$ .

$f'' (x) = \frac{d}{d x} (30 x^{4}) + \frac{d}{d x} (- 20 x^{3}) + \frac{d}{d x} (12 x^{2}) + \frac{d}{d x} (16 x)$

Evaluate $\frac{d}{d x} [30 x^{4}]$ .

Since $30$ is constant with respect to $x$ , the derivative of $30 x^{4}$ with respect to $x$ is $30 \frac{d}{d x} [x^{4}]$ .

$f'' (x) = 30 \frac{d}{d x} (x^{4}) + \frac{d}{d x} (- 20 x^{3}) + \frac{d}{d x} (12 x^{2}) + \frac{d}{d x} (16 x)$

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

$f'' (x) = 30 (4 x^{3}) + \frac{d}{d x} (- 20 x^{3}) + \frac{d}{d x} (12 x^{2}) + \frac{d}{d x} (16 x)$

Multiply $4$ by $30$ .

$f'' (x) = 120 x^{3} + \frac{d}{d x} (- 20 x^{3}) + \frac{d}{d x} (12 x^{2}) + \frac{d}{d x} (16 x)$

Evaluate $\frac{d}{d x} [- 20 x^{3}]$ .

Since $- 20$ is constant with respect to $x$ , the derivative of $- 20 x^{3}$ with respect to $x$ is $- 20 \frac{d}{d x} [x^{3}]$ .

$f'' (x) = 120 x^{3} - 20 \frac{d}{d x} x^{3} + \frac{d}{d x} (12 x^{2}) + \frac{d}{d x} (16 x)$

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

$f'' (x) = 120 x^{3} - 20 (3 x^{2}) + \frac{d}{d x} (12 x^{2}) + \frac{d}{d x} (16 x)$

Multiply $3$ by $- 20$ .

$f'' (x) = 120 x^{3} - 60 x^{2} + \frac{d}{d x} (12 x^{2}) + \frac{d}{d x} (16 x)$

Evaluate $\frac{d}{d x} [12 x^{2}]$ .

Since $12$ is constant with respect to $x$ , the derivative of $12 x^{2}$ with respect to $x$ is $12 \frac{d}{d x} [x^{2}]$ .

$f'' (x) = 120 x^{3} - 60 x^{2} + 12 \frac{d}{d x} (x^{2}) + \frac{d}{d x} (16 x)$

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

$f'' (x) = 120 x^{3} - 60 x^{2} + 12 (2 x) + \frac{d}{d x} (16 x)$

Multiply $2$ by $12$ .

$f'' (x) = 120 x^{3} - 60 x^{2} + 24 x + \frac{d}{d x} (16 x)$

Evaluate $\frac{d}{d x} [16 x]$ .

Since $16$ is constant with respect to $x$ , the derivative of $16 x$ with respect to $x$ is $16 \frac{d}{d x} [x]$ .

$f'' (x) = 120 x^{3} - 60 x^{2} + 24 x + 16 \frac{d}{d x} (x)$

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

$f'' (x) = 120 x^{3} - 60 x^{2} + 24 x + 16 \cdot 1$

Multiply $16$ by $1$ .

$f'' (x) = 120 x^{3} - 60 x^{2} + 24 x + 16$

Step 3

Find the third derivative.

By the Sum Rule, the derivative of $120 x^{3} - 60 x^{2} + 24 x + 16$ with respect to $x$ is $\frac{d}{d x} [120 x^{3}] + \frac{d}{d x} [- 60 x^{2}] + \frac{d}{d x} [24 x] + \frac{d}{d x} [16]$ .

$f''' (x) = \frac{d}{d x} (120 x^{3}) + \frac{d}{d x} (- 60 x^{2}) + \frac{d}{d x} (24 x) + \frac{d}{d x} (16)$

Evaluate $\frac{d}{d x} [120 x^{3}]$ .

Since $120$ is constant with respect to $x$ , the derivative of $120 x^{3}$ with respect to $x$ is $120 \frac{d}{d x} [x^{3}]$ .

$f''' (x) = 120 \frac{d}{d x} (x^{3}) + \frac{d}{d x} (- 60 x^{2}) + \frac{d}{d x} (24 x) + \frac{d}{d x} (16)$

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

$f''' (x) = 120 (3 x^{2}) + \frac{d}{d x} (- 60 x^{2}) + \frac{d}{d x} (24 x) + \frac{d}{d x} (16)$

Multiply $3$ by $120$ .

$f''' (x) = 360 x^{2} + \frac{d}{d x} (- 60 x^{2}) + \frac{d}{d x} (24 x) + \frac{d}{d x} (16)$

Evaluate $\frac{d}{d x} [- 60 x^{2}]$ .

Since $- 60$ is constant with respect to $x$ , the derivative of $- 60 x^{2}$ with respect to $x$ is $- 60 \frac{d}{d x} [x^{2}]$ .

$f''' (x) = 360 x^{2} - 60 \frac{d}{d x} x^{2} + \frac{d}{d x} (24 x) + \frac{d}{d x} (16)$

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

$f''' (x) = 360 x^{2} - 60 (2 x) + \frac{d}{d x} (24 x) + \frac{d}{d x} (16)$

Multiply $2$ by $- 60$ .

$f''' (x) = 360 x^{2} - 120 x + \frac{d}{d x} (24 x) + \frac{d}{d x} (16)$

Evaluate $\frac{d}{d x} [24 x]$ .

Since $24$ is constant with respect to $x$ , the derivative of $24 x$ with respect to $x$ is $24 \frac{d}{d x} [x]$ .

$f''' (x) = 360 x^{2} - 120 x + 24 \frac{d}{d x} (x) + \frac{d}{d x} (16)$

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

$f''' (x) = 360 x^{2} - 120 x + 24 \cdot 1 + \frac{d}{d x} (16)$

Multiply $24$ by $1$ .

$f''' (x) = 360 x^{2} - 120 x + 24 + \frac{d}{d x} (16)$

Differentiate using the Constant Rule.

Since $16$ is constant with respect to $x$ , the derivative of $16$ with respect to $x$ is $0$ .

$f''' (x) = 360 x^{2} - 120 x + 24 + 0$

Add $360 x^{2} - 120 x + 24$ and $0$ .

$f''' (x) = 360 x^{2} - 120 x + 24$

Step 4

Find the fourth derivative.

By the Sum Rule, the derivative of $360 x^{2} - 120 x + 24$ with respect to $x$ is $\frac{d}{d x} [360 x^{2}] + \frac{d}{d x} [- 120 x] + \frac{d}{d x} [24]$ .

$f^{4} (x) = \frac{d}{d x} (360 x^{2}) + \frac{d}{d x} (- 120 x) + \frac{d}{d x} (24)$

Evaluate $\frac{d}{d x} [360 x^{2}]$ .

Since $360$ is constant with respect to $x$ , the derivative of $360 x^{2}$ with respect to $x$ is $360 \frac{d}{d x} [x^{2}]$ .

$f^{4} (x) = 360 \frac{d}{d x} (x^{2}) + \frac{d}{d x} (- 120 x) + \frac{d}{d x} (24)$

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

$f^{4} (x) = 360 (2 x) + \frac{d}{d x} (- 120 x) + \frac{d}{d x} (24)$

Multiply $2$ by $360$ .

$f^{4} (x) = 720 x + \frac{d}{d x} (- 120 x) + \frac{d}{d x} (24)$

Evaluate $\frac{d}{d x} [- 120 x]$ .

Since $- 120$ is constant with respect to $x$ , the derivative of $- 120 x$ with respect to $x$ is $- 120 \frac{d}{d x} [x]$ .

$f^{4} (x) = 720 x - 120 \frac{d}{d x} x + \frac{d}{d x} (24)$

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

$f^{4} (x) = 720 x - 120 \cdot 1 + \frac{d}{d x} (24)$

Multiply $- 120$ by $1$ .

$f^{4} (x) = 720 x - 120 + \frac{d}{d x} (24)$

Differentiate using the Constant Rule.

Since $24$ is constant with respect to $x$ , the derivative of $24$ with respect to $x$ is $0$ .

$f^{4} (x) = 720 x - 120 + 0$

Add $720 x - 120$ and $0$ .

$f^{4} (x) = 720 x - 120$

Find the 4th Derivative 6x^5-5x^4+4x^3+8x^2+8

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