Find the Derivative – d/dx (x-1)(x^2+2)^3

$(x - 1) {(x^{2} + 2)}^{3}$

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 $f (x) = x - 1$ and $g (x) = {(x^{2} + 2)}^{3}$ .

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

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 $g (x) = x^{2} + 2$ .

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To apply the Chain Rule, set $u$ as $x^{2} + 2$ .

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

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

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

Replace all occurrences of $u$ with $x^{2} + 2$ .

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

Differentiate.

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Move $3$ to the left of $x - 1$ .

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

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

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

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

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

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

$3 (x - 1) {(x^{2} + 2)}^{2} (2 x + 0) + {(x^{2} + 2)}^{3} \frac{d}{d x} [x - 1]$

Simplify the expression.

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Add $2 x$ and $0$ .

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

Multiply $2$ by $3$ .

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

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

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

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

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

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

$6 (x - 1) {(x^{2} + 2)}^{2} x + {(x^{2} + 2)}^{3} (1 + 0)$

Simplify the expression.

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Add $1$ and $0$ .

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

Multiply ${(x^{2} + 2)}^{3}$ by $1$ .

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

Simplify.

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Apply the distributive property.

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

Multiply $6$ by $- 1$ .

$(6 x - 6) {(x^{2} + 2)}^{2} x + {(x^{2} + 2)}^{3}$

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

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

${(x^{2} + 2)}^{2} ((6 x - 6) x) + {(x^{2} + 2)}^{3}$

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

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

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

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

Find the Derivative – d/dx (x-1)(x^2+2)^3

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