Find the Derivative - d/dx natural log of -x^3+6x^2+96x+1

$\ln (- x^{3} + 6 x^{2} + 96 x + 1)$

Step 1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = - x^{3} + 6 x^{2} + 96 x + 1$ .

To apply the Chain Rule, set $u$ as $- x^{3} + 6 x^{2} + 96 x + 1$ .

$\frac{d}{d u} [\ln (u)] \frac{d}{d x} [- x^{3} + 6 x^{2} + 96 x + 1]$

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

$\frac{1}{u} \frac{d}{d x} [- x^{3} + 6 x^{2} + 96 x + 1]$

Replace all occurrences of $u$ with $- x^{3} + 6 x^{2} + 96 x + 1$ .

$\frac{1}{- x^{3} + 6 x^{2} + 96 x + 1} \frac{d}{d x} [- x^{3} + 6 x^{2} + 96 x + 1]$

Step 2

Differentiate.

By the Sum Rule, the derivative of $- x^{3} + 6 x^{2} + 96 x + 1$ with respect to $x$ is $\frac{d}{d x} [- x^{3}] + \frac{d}{d x} [6 x^{2}] + \frac{d}{d x} [96 x] + \frac{d}{d x} [1]$ .

$\frac{1}{- x^{3} + 6 x^{2} + 96 x + 1} (\frac{d}{d x} [- x^{3}] + \frac{d}{d x} [6 x^{2}] + \frac{d}{d x} [96 x] + \frac{d}{d x} [1])$

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

$\frac{1}{- x^{3} + 6 x^{2} + 96 x + 1} (- \frac{d}{d x} [x^{3}] + \frac{d}{d x} [6 x^{2}] + \frac{d}{d x} [96 x] + \frac{d}{d x} [1])$

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

$\frac{1}{- x^{3} + 6 x^{2} + 96 x + 1} (- (3 x^{2}) + \frac{d}{d x} [6 x^{2}] + \frac{d}{d x} [96 x] + \frac{d}{d x} [1])$

Multiply $3$ by $- 1$ .

$\frac{1}{- x^{3} + 6 x^{2} + 96 x + 1} (- 3 x^{2} + \frac{d}{d x} [6 x^{2}] + \frac{d}{d x} [96 x] + \frac{d}{d x} [1])$

Since $6$ is constant with respect to $x$ , the derivative of $6 x^{2}$ with respect to $x$ is $6 \frac{d}{d x} [x^{2}]$ .

$\frac{1}{- x^{3} + 6 x^{2} + 96 x + 1} (- 3 x^{2} + 6 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [96 x] + \frac{d}{d x} [1])$

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

$\frac{1}{- x^{3} + 6 x^{2} + 96 x + 1} (- 3 x^{2} + 6 (2 x) + \frac{d}{d x} [96 x] + \frac{d}{d x} [1])$

Multiply $2$ by $6$ .

$\frac{1}{- x^{3} + 6 x^{2} + 96 x + 1} (- 3 x^{2} + 12 x + \frac{d}{d x} [96 x] + \frac{d}{d x} [1])$

Since $96$ is constant with respect to $x$ , the derivative of $96 x$ with respect to $x$ is $96 \frac{d}{d x} [x]$ .

$\frac{1}{- x^{3} + 6 x^{2} + 96 x + 1} (- 3 x^{2} + 12 x + 96 \frac{d}{d x} [x] + \frac{d}{d x} [1])$

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

$\frac{1}{- x^{3} + 6 x^{2} + 96 x + 1} (- 3 x^{2} + 12 x + 96 \cdot 1 + \frac{d}{d x} [1])$

Multiply $96$ by $1$ .

$\frac{1}{- x^{3} + 6 x^{2} + 96 x + 1} (- 3 x^{2} + 12 x + 96 + \frac{d}{d x} [1])$

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

$\frac{1}{- x^{3} + 6 x^{2} + 96 x + 1} (- 3 x^{2} + 12 x + 96 + 0)$

Add $- 3 x^{2} + 12 x + 96$ and $0$ .

$\frac{1}{- x^{3} + 6 x^{2} + 96 x + 1} (- 3 x^{2} + 12 x + 96)$

Step 3

Simplify.

Reorder the factors of $\frac{1}{- x^{3} + 6 x^{2} + 96 x + 1} (- 3 x^{2} + 12 x + 96)$ .

$(- 3 x^{2} + 12 x + 96) \frac{1}{- x^{3} + 6 x^{2} + 96 x + 1}$

Multiply $- 3 x^{2} + 12 x + 96$ by $\frac{1}{- x^{3} + 6 x^{2} + 96 x + 1}$ .

$\frac{- 3 x^{2} + 12 x + 96}{- x^{3} + 6 x^{2} + 96 x + 1}$

Simplify the numerator.

Factor $3$ out of $- 3 x^{2} + 12 x + 96$ .

Factor $3$ out of $- 3 x^{2}$ .

$\frac{3 (- x^{2}) + 12 x + 96}{- x^{3} + 6 x^{2} + 96 x + 1}$

Factor $3$ out of $12 x$ .

$\frac{3 (- x^{2}) + 3 (4 x) + 96}{- x^{3} + 6 x^{2} + 96 x + 1}$

Factor $3$ out of $96$ .

$\frac{3 (- x^{2}) + 3 (4 x) + 3 (32)}{- x^{3} + 6 x^{2} + 96 x + 1}$

Factor $3$ out of $3 (- x^{2}) + 3 (4 x)$ .

$\frac{3 (- x^{2} + 4 x) + 3 (32)}{- x^{3} + 6 x^{2} + 96 x + 1}$

Factor $3$ out of $3 (- x^{2} + 4 x) + 3 (32)$ .

$\frac{3 (- x^{2} + 4 x + 32)}{- x^{3} + 6 x^{2} + 96 x + 1}$

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 1 \cdot 32 = - 32$ and whose sum is $b = 4$ .

Factor $4$ out of $4 x$ .

$\frac{3 (- x^{2} + 4 (x) + 32)}{- x^{3} + 6 x^{2} + 96 x + 1}$

Rewrite $4$ as $- 4$ plus $8$

$\frac{3 (- x^{2} + (- 4 + 8) x + 32)}{- x^{3} + 6 x^{2} + 96 x + 1}$

Apply the distributive property.

$\frac{3 (- x^{2} - 4 x + 8 x + 32)}{- x^{3} + 6 x^{2} + 96 x + 1}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{3 ((- x^{2} - 4 x) + 8 x + 32)}{- x^{3} + 6 x^{2} + 96 x + 1}$

Factor out the greatest common factor (GCF) from each group.

$\frac{3 (x (- x - 4) - 8 (- x - 4))}{- x^{3} + 6 x^{2} + 96 x + 1}$

Factor the polynomial by factoring out the greatest common factor, $- x - 4$ .

$\frac{3 ((- x - 4) (x - 8))}{- x^{3} + 6 x^{2} + 96 x + 1}$

$\frac{3 (- x - 4) (x - 8)}{- x^{3} + 6 x^{2} + 96 x + 1}$

Factor $- 1$ out of $- x$ .

$\frac{3 (- (x) - 4) (x - 8)}{- x^{3} + 6 x^{2} + 96 x + 1}$

Rewrite $- 4$ as $- 1 (4)$ .

$\frac{3 (- (x) - 1 (4)) (x - 8)}{- x^{3} + 6 x^{2} + 96 x + 1}$

Factor $- 1$ out of $- (x) - 1 (4)$ .

$\frac{3 (- (x + 4)) (x - 8)}{- x^{3} + 6 x^{2} + 96 x + 1}$

Rewrite $- (x + 4)$ as $- 1 (x + 4)$ .

$\frac{3 (- 1 (x + 4)) (x - 8)}{- x^{3} + 6 x^{2} + 96 x + 1}$

Factor $- 1$ out of $- x^{3}$ .

$\frac{3 (- 1 (x + 4)) (x - 8)}{- (x^{3}) + 6 x^{2} + 96 x + 1}$

Factor $- 1$ out of $6 x^{2}$ .

$\frac{3 (- 1 (x + 4)) (x - 8)}{- (x^{3}) - (- 6 x^{2}) + 96 x + 1}$

Factor $- 1$ out of $- (x^{3}) - (- 6 x^{2})$ .

$\frac{3 (- 1 (x + 4)) (x - 8)}{- (x^{3} - 6 x^{2}) + 96 x + 1}$

Factor $- 1$ out of $96 x$ .

$\frac{3 (- 1 (x + 4)) (x - 8)}{- (x^{3} - 6 x^{2}) - (- 96 x) + 1}$

Factor $- 1$ out of $- (x^{3} - 6 x^{2}) - (- 96 x)$ .

$\frac{3 (- 1 (x + 4)) (x - 8)}{- (x^{3} - 6 x^{2} - 96 x) + 1}$

Rewrite $1$ as $- 1 (- 1)$ .

$\frac{3 (- 1 (x + 4)) (x - 8)}{- (x^{3} - 6 x^{2} - 96 x) - 1 (- 1)}$

Factor $- 1$ out of $- (x^{3} - 6 x^{2} - 96 x) - 1 (- 1)$ .

$\frac{3 (- 1 (x + 4)) (x - 8)}{- (x^{3} - 6 x^{2} - 96 x - 1)}$

Rewrite $- (x^{3} - 6 x^{2} - 96 x - 1)$ as $- 1 (x^{3} - 6 x^{2} - 96 x - 1)$ .

$\frac{3 (- 1 (x + 4)) (x - 8)}{- 1 (x^{3} - 6 x^{2} - 96 x - 1)}$

Cancel the common factor.

$\frac{3 (- 1 (x + 4)) (x - 8)}{- 1 (x^{3} - 6 x^{2} - 96 x - 1)}$

Rewrite the expression.

$\frac{3 (x + 4) (x - 8)}{x^{3} - 6 x^{2} - 96 x - 1}$

Find the Derivative - d/dx natural log of -x^3+6x^2+96x+1

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