Math / Calculus

QuestionFind $y^{\prime}$ by (a) applying the Product Rule and (b) multiplying the factors to produce a sum of simpler terms to differentiate. $y=\left(3-x^{2}\right)\left(x^{3}-4 x+2\right)$ a. Apply the Product Rule. Let $u=\left(3-x^{2}\right)$ and $v=\left(x^{3}-4 x+2\right)$ . $\frac{d}{d x}(u v)=\left(3-x^{2}\right)\left(3 x^{2}-4\right)+\left(x^{3}-4 x+2\right)(-2 x)$ b. Multiply the factors of the original expression, $u$ and $v$ , to produce a sum of simpler terms. $\mathrm{y}=\square$ (Simplify your answer.)

Studdy Solution

STEP 1

1. We are given the function $y = (3-x^2)(x^3-4x+2)$ .
2. We need to find $y'$ using two methods: (a) the Product Rule and (b) by expanding the expression and differentiating.

STEP 2

1. Apply the Product Rule to find $y'$ .
2. Expand the expression $y = (3-x^2)(x^3-4x+2)$ into a sum of simpler terms.
3. Differentiate the expanded expression to find $y'$ .

STEP 3

Identify the functions $u$ and $v$ :
Let $u = 3-x^2$ and $v = x^3-4x+2$ .

STEP 4

Differentiate $u$ and $v$ :
$u' = \frac{d}{dx}(3-x^2) = -2x$ $v' = \frac{d}{dx}(x^3-4x+2) = 3x^2 - 4$

STEP 5

Apply the Product Rule:
$y' = u'v + uv'$ $y' = (-2x)(x^3-4x+2) + (3-x^2)(3x^2-4)$

STEP 6

Simplify the expression:
$y' = (-2x)(x^3) + (-2x)(-4x) + (-2x)(2) + (3)(3x^2) + (-x^2)(3x^2) + (3)(-4) + (-x^2)(-4)$
$y' = -2x^4 + 8x^2 - 4x + 9x^2 - 3x^4 - 12 + 4x^2$
Combine like terms:
$y' = (-2x^4 - 3x^4) + (8x^2 + 9x^2 + 4x^2) - 4x - 12$
$y' = -5x^4 + 21x^2 - 4x - 12$

STEP 7

Expand the original expression $y = (3-x^2)(x^3-4x+2)$ :
$y = 3(x^3) + 3(-4x) + 3(2) + (-x^2)(x^3) + (-x^2)(-4x) + (-x^2)(2)$
$y = 3x^3 - 12x + 6 - x^5 + 4x^3 - 2x^2$
Combine like terms:
$y = -x^5 + (3x^3 + 4x^3) - 2x^2 - 12x + 6$
$y = -x^5 + 7x^3 - 2x^2 - 12x + 6$

STEP 8

Differentiate the expanded expression:
$y' = \frac{d}{dx}(-x^5 + 7x^3 - 2x^2 - 12x + 6)$
$y' = -5x^4 + 21x^2 - 4x - 12$
The derivative $y'$ is:
$\boxed{-5x^4 + 21x^2 - 4x - 12}$

Was this helpful?

Studdy Solution

STEP 1

1. We are given the function $y = (3-x^2)(x^3-4x+2)$ .
2. We need to find $y'$ using two methods: (a) the Product Rule and (b) by expanding the expression and differentiating.

STEP 2

1. Apply the Product Rule to find $y'$ .
2. Expand the expression $y = (3-x^2)(x^3-4x+2)$ into a sum of simpler terms.
3. Differentiate the expanded expression to find $y'$ .

STEP 3

Identify the functions $u$ and $v$ :
Let $u = 3-x^2$ and $v = x^3-4x+2$ .

STEP 4

Differentiate $u$ and $v$ :
$u' = \frac{d}{dx}(3-x^2) = -2x$ $v' = \frac{d}{dx}(x^3-4x+2) = 3x^2 - 4$

STEP 5

Apply the Product Rule:
$y' = u'v + uv'$ $y' = (-2x)(x^3-4x+2) + (3-x^2)(3x^2-4)$

STEP 6

Simplify the expression:
$y' = (-2x)(x^3) + (-2x)(-4x) + (-2x)(2) + (3)(3x^2) + (-x^2)(3x^2) + (3)(-4) + (-x^2)(-4)$
$y' = -2x^4 + 8x^2 - 4x + 9x^2 - 3x^4 - 12 + 4x^2$
Combine like terms:
$y' = (-2x^4 - 3x^4) + (8x^2 + 9x^2 + 4x^2) - 4x - 12$
$y' = -5x^4 + 21x^2 - 4x - 12$

STEP 7

Expand the original expression $y = (3-x^2)(x^3-4x+2)$ :
$y = 3(x^3) + 3(-4x) + 3(2) + (-x^2)(x^3) + (-x^2)(-4x) + (-x^2)(2)$
$y = 3x^3 - 12x + 6 - x^5 + 4x^3 - 2x^2$
Combine like terms:
$y = -x^5 + (3x^3 + 4x^3) - 2x^2 - 12x + 6$
$y = -x^5 + 7x^3 - 2x^2 - 12x + 6$

STEP 8

Differentiate the expanded expression:
$y' = \frac{d}{dx}(-x^5 + 7x^3 - 2x^2 - 12x + 6)$
$y' = -5x^4 + 21x^2 - 4x - 12$
The derivative $y'$ is:
$\boxed{-5x^4 + 21x^2 - 4x - 12}$

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Studdy Solution

STEP 1

1. We are given the function y=(3−x2)(x3−4x+2) y = (3-x^2)(x^3-4x+2) y=(3−x2)(x3−4x+2).2. We need to find y′ y' y′ using two methods: (a) the Product Rule and (b) by expanding the expression and differentiating.

STEP 2

1. Apply the Product Rule to find y′ y' y′.2. Expand the expression y=(3−x2)(x3−4x+2) y = (3-x^2)(x^3-4x+2) y=(3−x2)(x3−4x+2) into a sum of simpler terms.3. Differentiate the expanded expression to find y′ y' y′.

STEP 3

Identify the functions u u u and v v v:Let u=3−x2 u = 3-x^2 u=3−x2 and v=x3−4x+2 v = x^3-4x+2 v=x3−4x+2.

STEP 4

Differentiate u u u and v v v:u′=ddx(3−x2)=−2x u' = \frac{d}{dx}(3-x^2) = -2x u′=dxd​(3−x2)=−2x v′=ddx(x3−4x+2)=3x2−4 v' = \frac{d}{dx}(x^3-4x+2) = 3x^2 - 4 v′=dxd​(x3−4x+2)=3x2−4

STEP 5

Apply the Product Rule:y′=u′v+uv′ y' = u'v + uv' y′=u′v+uv′ y′=(−2x)(x3−4x+2)+(3−x2)(3x2−4) y' = (-2x)(x^3-4x+2) + (3-x^2)(3x^2-4) y′=(−2x)(x3−4x+2)+(3−x2)(3x2−4)

STEP 6

STEP 7

STEP 8

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. We are given the function $y = (3-x^2)(x^3-4x+2)$ .
2. We need to find $y'$ using two methods: (a) the Product Rule and (b) by expanding the expression and differentiating.

1. Apply the Product Rule to find $y'$ .
2. Expand the expression $y = (3-x^2)(x^3-4x+2)$ into a sum of simpler terms.
3. Differentiate the expanded expression to find $y'$ .

Identify the functions $u$ and $v$ :
Let $u = 3-x^2$ and $v = x^3-4x+2$ .

Differentiate $u$ and $v$ :
$u' = \frac{d}{dx}(3-x^2) = -2x$ $v' = \frac{d}{dx}(x^3-4x+2) = 3x^2 - 4$

Apply the Product Rule:
$y' = u'v + uv'$ $y' = (-2x)(x^3-4x+2) + (3-x^2)(3x^2-4)$