Solved on Feb 05, 2024

Find the derivative of $f(x) = 5x^4(x^3 - 2)$ and identify the correct application of the product rule.

STEP 1

Assumptions
1. The function $f(x)$ is given by $f(x)=5x^{4}(x^{3}-2)$ .
2. We need to find the derivative $f'(x)$ using the product rule.
3. The product rule states that if $u(x)$ and $v(x)$ are differentiable functions, then the derivative of their product $u(x)v(x)$ is given by $u'(x)v(x) + u(x)v'(x)$ .

STEP 2

Identify the two parts of the function that are being multiplied, which we will call $u(x)$ and $v(x)$ .
$u(x) = 5x^{4} \quad \text{and} \quad v(x) = x^{3}-2$

STEP 3

Find the derivative of $u(x)$ with respect to $x$ .
$u'(x) = \frac{d}{dx}(5x^{4})$

STEP 4

Apply the power rule to find $u'(x)$ .
$u'(x) = 5 \cdot 4x^{4-1}$

STEP 5

Simplify the expression for $u'(x)$ .
$u'(x) = 20x^{3}$

STEP 6

Find the derivative of $v(x)$ with respect to $x$ .
$v'(x) = \frac{d}{dx}(x^{3}-2)$

STEP 7

Apply the power rule to the $x^{3}$ term and note that the derivative of a constant is zero.
$v'(x) = 3x^{3-1} - 0$

STEP 8

Simplify the expression for $v'(x)$ .
$v'(x) = 3x^{2}$

STEP 9

Apply the product rule to find $f'(x)$ .
$f'(x) = u'(x)v(x) + u(x)v'(x)$

STEP 10

Substitute the expressions for $u'(x)$ , $v(x)$ , $u(x)$ , and $v'(x)$ into the product rule formula.
$f'(x) = (20x^{3})(x^{3}-2) + (5x^{4})(3x^{2})$

STEP 11

Distribute $20x^{3}$ across the terms in the parentheses $(x^{3}-2)$ .
$f'(x) = 20x^{3} \cdot x^{3} - 20x^{3} \cdot 2 + (5x^{4})(3x^{2})$

STEP 12

Simplify the distributed terms.
$f'(x) = 20x^{6} - 40x^{3} + (5x^{4})(3x^{2})$

STEP 13

Multiply $(5x^{4})$ by $(3x^{2})$ .
$f'(x) = 20x^{6} - 40x^{3} + 15x^{4} \cdot x^{2}$

STEP 14

Simplify the last term by adding the exponents.
$f'(x) = 20x^{6} - 40x^{3} + 15x^{6}$

STEP 15

Combine like terms.
$f'(x) = (20x^{6} + 15x^{6}) - 40x^{3}$

STEP 16

Add the coefficients of the $x^{6}$ terms.
$f'(x) = 35x^{6} - 40x^{3}$
Now, let's identify which of the given options corresponds to the correct application of the product rule.

STEP 17

Option A is incorrect because it does not apply the product rule correctly; it only multiplies parts of the derivatives without considering both $u(x)$ and $v(x)$ .

STEP 18

Option B is incorrect because it subtracts the second term instead of adding it, which is not consistent with the product rule.

STEP 19

Option C is incorrect because it multiplies $5x^{4}$ by $20x^{3}$ , which is not the correct application of the product rule.

STEP 20

Option D correctly applies the product rule, showing the sum of the product of the derivative of the first function and the second function, and the product of the first function and the derivative of the second function.
Therefore, the correct application of the product rule and the derivative $f'(x)$ is given by Option D:
$f'(x) = 35x^{6} - 40x^{3}$

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Find the derivative of f(x)=5x4(x3−2)f(x) = 5x^4(x^3 - 2)f(x)=5x4(x3−2) and identify the correct application of the product rule.

STEP 1

STEP 2

Identify the two parts of the function that are being multiplied, which we will call u(x)u(x)u(x) and v(x)v(x)v(x).u(x)=5x4andv(x)=x3−2u(x) = 5x^{4} \quad \text{and} \quad v(x) = x^{3}-2u(x)=5x4andv(x)=x3−2

STEP 3

Find the derivative of u(x)u(x)u(x) with respect to xxx.u′(x)=ddx(5x4)u'(x) = \frac{d}{dx}(5x^{4})u′(x)=dxd​(5x4)

STEP 4

Apply the power rule to find u′(x)u'(x)u′(x).u′(x)=5⋅4x4−1u'(x) = 5 \cdot 4x^{4-1}u′(x)=5⋅4x4−1

STEP 5

Simplify the expression for u′(x)u'(x)u′(x).u′(x)=20x3u'(x) = 20x^{3}u′(x)=20x3

STEP 6

Find the derivative of v(x)v(x)v(x) with respect to xxx.v′(x)=ddx(x3−2)v'(x) = \frac{d}{dx}(x^{3}-2)v′(x)=dxd​(x3−2)

STEP 7

Apply the power rule to the x3x^{3}x3 term and note that the derivative of a constant is zero.v′(x)=3x3−1−0v'(x) = 3x^{3-1} - 0v′(x)=3x3−1−0

STEP 8

Simplify the expression for v′(x)v'(x)v′(x).v′(x)=3x2v'(x) = 3x^{2}v′(x)=3x2

STEP 9

Apply the product rule to find f′(x)f'(x)f′(x).f′(x)=u′(x)v(x)+u(x)v′(x)f'(x) = u'(x)v(x) + u(x)v'(x)f′(x)=u′(x)v(x)+u(x)v′(x)

STEP 10

Substitute the expressions for u′(x)u'(x)u′(x), v(x)v(x)v(x), u(x)u(x)u(x), and v′(x)v'(x)v′(x) into the product rule formula.f′(x)=(20x3)(x3−2)+(5x4)(3x2)f'(x) = (20x^{3})(x^{3}-2) + (5x^{4})(3x^{2})f′(x)=(20x3)(x3−2)+(5x4)(3x2)

STEP 11

Distribute 20x320x^{3}20x3 across the terms in the parentheses (x3−2)(x^{3}-2)(x3−2).f′(x)=20x3⋅x3−20x3⋅2+(5x4)(3x2)f'(x) = 20x^{3} \cdot x^{3} - 20x^{3} \cdot 2 + (5x^{4})(3x^{2})f′(x)=20x3⋅x3−20x3⋅2+(5x4)(3x2)

STEP 12

Simplify the distributed terms.f′(x)=20x6−40x3+(5x4)(3x2)f'(x) = 20x^{6} - 40x^{3} + (5x^{4})(3x^{2})f′(x)=20x6−40x3+(5x4)(3x2)

STEP 13

Multiply (5x4)(5x^{4})(5x4) by (3x2)(3x^{2})(3x2).f′(x)=20x6−40x3+15x4⋅x2f'(x) = 20x^{6} - 40x^{3} + 15x^{4} \cdot x^{2}f′(x)=20x6−40x3+15x4⋅x2

STEP 14

Simplify the last term by adding the exponents.f′(x)=20x6−40x3+15x6f'(x) = 20x^{6} - 40x^{3} + 15x^{6}f′(x)=20x6−40x3+15x6

STEP 15

Combine like terms.f′(x)=(20x6+15x6)−40x3f'(x) = (20x^{6} + 15x^{6}) - 40x^{3}f′(x)=(20x6+15x6)−40x3

STEP 16

Add the coefficients of the x6x^{6}x6 terms.f′(x)=35x6−40x3f'(x) = 35x^{6} - 40x^{3}f′(x)=35x6−40x3Now, let's identify which of the given options corresponds to the correct application of the product rule.

STEP 17

Option A is incorrect because it does not apply the product rule correctly; it only multiplies parts of the derivatives without considering both u(x)u(x)u(x) and v(x)v(x)v(x).

STEP 18

Option B is incorrect because it subtracts the second term instead of adding it, which is not consistent with the product rule.

STEP 19

Option C is incorrect because it multiplies 5x45x^{4}5x4 by 20x320x^{3}20x3, which is not the correct application of the product rule.

STEP 20

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Find the derivative of $f(x) = 5x^4(x^3 - 2)$ and identify the correct application of the product rule.

Identify the two parts of the function that are being multiplied, which we will call $u(x)$ and $v(x)$ .
$u(x) = 5x^{4} \quad \text{and} \quad v(x) = x^{3}-2$

Find the derivative of $u(x)$ with respect to $x$ .
$u'(x) = \frac{d}{dx}(5x^{4})$

Apply the power rule to find $u'(x)$ .
$u'(x) = 5 \cdot 4x^{4-1}$

Simplify the expression for $u'(x)$ .
$u'(x) = 20x^{3}$

Find the derivative of $v(x)$ with respect to $x$ .
$v'(x) = \frac{d}{dx}(x^{3}-2)$

Apply the power rule to the $x^{3}$ term and note that the derivative of a constant is zero.
$v'(x) = 3x^{3-1} - 0$

Simplify the expression for $v'(x)$ .
$v'(x) = 3x^{2}$

Apply the product rule to find $f'(x)$ .
$f'(x) = u'(x)v(x) + u(x)v'(x)$

Substitute the expressions for $u'(x)$ , $v(x)$ , $u(x)$ , and $v'(x)$ into the product rule formula.
$f'(x) = (20x^{3})(x^{3}-2) + (5x^{4})(3x^{2})$

Distribute $20x^{3}$ across the terms in the parentheses $(x^{3}-2)$ .
$f'(x) = 20x^{3} \cdot x^{3} - 20x^{3} \cdot 2 + (5x^{4})(3x^{2})$

Simplify the distributed terms.
$f'(x) = 20x^{6} - 40x^{3} + (5x^{4})(3x^{2})$

Multiply $(5x^{4})$ by $(3x^{2})$ .
$f'(x) = 20x^{6} - 40x^{3} + 15x^{4} \cdot x^{2}$

Simplify the last term by adding the exponents.
$f'(x) = 20x^{6} - 40x^{3} + 15x^{6}$

Combine like terms.
$f'(x) = (20x^{6} + 15x^{6}) - 40x^{3}$

Add the coefficients of the $x^{6}$ terms.
$f'(x) = 35x^{6} - 40x^{3}$
Now, let's identify which of the given options corresponds to the correct application of the product rule.

Option A is incorrect because it does not apply the product rule correctly; it only multiplies parts of the derivatives without considering both $u(x)$ and $v(x)$ .

Option C is incorrect because it multiplies $5x^{4}$ by $20x^{3}$ , which is not the correct application of the product rule.