Math / Calculus

QuestionQuestion
Given the function $y=3 x^{2} \sin ^{4}(x)$ , find $\frac{d y}{d x}$ in any form. Note: make sure to write trig functions using parentheses like $\sin ^{2}(x)$ instea interpreter might not be able to understand your expression.

Studdy Solution

STEP 1

1. The function $y = 3x^2 \sin^4(x)$ involves both polynomial and trigonometric components.
2. To find the derivative $\frac{d y}{d x}$ , we will need to use the product rule and the chain rule.

STEP 2

1. Apply the product rule to differentiate the function $y = 3x^2 \sin^4(x)$ .
2. Use the chain rule to differentiate $\sin^4(x)$ .

STEP 3

Identify the two functions involved in the product rule. Let $u = 3x^2$ and $v = \sin^4(x)$ .
$y = u \cdot v = 3x^2 \sin^4(x)$

STEP 4

Recall the product rule for differentiation:
$\frac{d}{dx} (u \cdot v) = u' \cdot v + u \cdot v'$

STEP 5

Differentiate $u = 3x^2$ with respect to $x$ :
$u' = \frac{d}{dx} (3x^2) = 6x$

STEP 6

Differentiate $v = \sin^4(x)$ with respect to $x$ using the chain rule. Let $w = \sin(x)$ , so $v = w^4$ .
First, differentiate $w^4$ with respect to $w$ :
$\frac{d}{dw} (w^4) = 4w^3$

STEP 7

Now, differentiate $w = \sin(x)$ with respect to $x$ :
$\frac{d}{dx} (\sin(x)) = \cos(x)$

STEP 8

Apply the chain rule to combine these derivatives:
$v' = \frac{d}{dx} (\sin^4(x)) = 4 \sin^3(x) \cdot \cos(x)$

STEP 9

Substitute $u$ , $u'$ , $v$ , and $v'$ into the product rule formula:
$\frac{d}{dx} (3x^2 \sin^4(x)) = u' \cdot v + u \cdot v'$ $= 6x \cdot \sin^4(x) + 3x^2 \cdot 4 \sin^3(x) \cos(x)$

STEP 10

Simplify the expression:
$= 6x \cdot \sin^4(x) + 12x^2 \sin^3(x) \cos(x)$

STEP 11

Write the final derivative:
$\frac{d y}{d x} = 6x \sin^4(x) + 12x^2 \sin^3(x) \cos(x)$

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QuestionQuestion
Given the function $y=3 x^{2} \sin ^{4}(x)$ , find $\frac{d y}{d x}$ in any form. Note: make sure to write trig functions using parentheses like $\sin ^{2}(x)$ instea interpreter might not be able to understand your expression.

Studdy Solution

STEP 1

1. The function $y = 3x^2 \sin^4(x)$ involves both polynomial and trigonometric components.
2. To find the derivative $\frac{d y}{d x}$ , we will need to use the product rule and the chain rule.

STEP 2

1. Apply the product rule to differentiate the function $y = 3x^2 \sin^4(x)$ .
2. Use the chain rule to differentiate $\sin^4(x)$ .

STEP 3

Identify the two functions involved in the product rule. Let $u = 3x^2$ and $v = \sin^4(x)$ .
$y = u \cdot v = 3x^2 \sin^4(x)$

STEP 4

Recall the product rule for differentiation:
$\frac{d}{dx} (u \cdot v) = u' \cdot v + u \cdot v'$

STEP 5

Differentiate $u = 3x^2$ with respect to $x$ :
$u' = \frac{d}{dx} (3x^2) = 6x$

STEP 6

Differentiate $v = \sin^4(x)$ with respect to $x$ using the chain rule. Let $w = \sin(x)$ , so $v = w^4$ .
First, differentiate $w^4$ with respect to $w$ :
$\frac{d}{dw} (w^4) = 4w^3$

STEP 7

Now, differentiate $w = \sin(x)$ with respect to $x$ :
$\frac{d}{dx} (\sin(x)) = \cos(x)$

STEP 8

Apply the chain rule to combine these derivatives:
$v' = \frac{d}{dx} (\sin^4(x)) = 4 \sin^3(x) \cdot \cos(x)$

STEP 9

Substitute $u$ , $u'$ , $v$ , and $v'$ into the product rule formula:
$\frac{d}{dx} (3x^2 \sin^4(x)) = u' \cdot v + u \cdot v'$ $= 6x \cdot \sin^4(x) + 3x^2 \cdot 4 \sin^3(x) \cos(x)$

STEP 10

Simplify the expression:
$= 6x \cdot \sin^4(x) + 12x^2 \sin^3(x) \cos(x)$

STEP 11

Write the final derivative:
$\frac{d y}{d x} = 6x \sin^4(x) + 12x^2 \sin^3(x) \cos(x)$

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QuestionQuestionGiven the function y=3x2sin⁡4(x)y=3 x^{2} \sin ^{4}(x)y=3x2sin4(x), find dydx\frac{d y}{d x}dxdy​ in any form. Note: make sure to write trig functions using parentheses like sin⁡2(x)\sin ^{2}(x)sin2(x) instea interpreter might not be able to understand your expression.

Studdy Solution

STEP 1

1. The function y=3x2sin⁡4(x) y = 3x^2 \sin^4(x) y=3x2sin4(x) involves both polynomial and trigonometric components.2. To find the derivative dydx\frac{d y}{d x}dxdy​, we will need to use the product rule and the chain rule.

STEP 2

1. Apply the product rule to differentiate the function y=3x2sin⁡4(x) y = 3x^2 \sin^4(x) y=3x2sin4(x).2. Use the chain rule to differentiate sin⁡4(x)\sin^4(x)sin4(x).

STEP 3

Identify the two functions involved in the product rule. Let u=3x2 u = 3x^2 u=3x2 and v=sin⁡4(x) v = \sin^4(x) v=sin4(x).y=u⋅v=3x2sin⁡4(x) y = u \cdot v = 3x^2 \sin^4(x) y=u⋅v=3x2sin4(x)

STEP 4

Recall the product rule for differentiation: ddx(u⋅v)=u′⋅v+u⋅v′ \frac{d}{dx} (u \cdot v) = u' \cdot v + u \cdot v' dxd​(u⋅v)=u′⋅v+u⋅v′

STEP 5

Differentiate u=3x2 u = 3x^2 u=3x2 with respect to x x x:u′=ddx(3x2)=6x u' = \frac{d}{dx} (3x^2) = 6x u′=dxd​(3x2)=6x

STEP 6

Differentiate v=sin⁡4(x) v = \sin^4(x) v=sin4(x) with respect to x x x using the chain rule. Let w=sin⁡(x) w = \sin(x) w=sin(x), so v=w4 v = w^4 v=w4.First, differentiate w4 w^4 w4 with respect to w w w:ddw(w4)=4w3 \frac{d}{dw} (w^4) = 4w^3 dwd​(w4)=4w3

STEP 7

Now, differentiate w=sin⁡(x) w = \sin(x) w=sin(x) with respect to x x x:ddx(sin⁡(x))=cos⁡(x) \frac{d}{dx} (\sin(x)) = \cos(x) dxd​(sin(x))=cos(x)

STEP 8

Apply the chain rule to combine these derivatives:v′=ddx(sin⁡4(x))=4sin⁡3(x)⋅cos⁡(x) v' = \frac{d}{dx} (\sin^4(x)) = 4 \sin^3(x) \cdot \cos(x) v′=dxd​(sin4(x))=4sin3(x)⋅cos(x)

STEP 9

STEP 10

Simplify the expression:=6x⋅sin⁡4(x)+12x2sin⁡3(x)cos⁡(x) = 6x \cdot \sin^4(x) + 12x^2 \sin^3(x) \cos(x) =6x⋅sin4(x)+12x2sin3(x)cos(x)

STEP 11

Write the final derivative:dydx=6xsin⁡4(x)+12x2sin⁡3(x)cos⁡(x) \frac{d y}{d x} = 6x \sin^4(x) + 12x^2 \sin^3(x) \cos(x) dxdy​=6xsin4(x)+12x2sin3(x)cos(x)

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QuestionQuestion
Given the function $y=3 x^{2} \sin ^{4}(x)$ , find $\frac{d y}{d x}$ in any form. Note: make sure to write trig functions using parentheses like $\sin ^{2}(x)$ instea interpreter might not be able to understand your expression.

1. The function $y = 3x^2 \sin^4(x)$ involves both polynomial and trigonometric components.
2. To find the derivative $\frac{d y}{d x}$ , we will need to use the product rule and the chain rule.

1. Apply the product rule to differentiate the function $y = 3x^2 \sin^4(x)$ .
2. Use the chain rule to differentiate $\sin^4(x)$ .

Identify the two functions involved in the product rule. Let $u = 3x^2$ and $v = \sin^4(x)$ .
$y = u \cdot v = 3x^2 \sin^4(x)$

Recall the product rule for differentiation:
$\frac{d}{dx} (u \cdot v) = u' \cdot v + u \cdot v'$

Differentiate $u = 3x^2$ with respect to $x$ :
$u' = \frac{d}{dx} (3x^2) = 6x$

Differentiate $v = \sin^4(x)$ with respect to $x$ using the chain rule. Let $w = \sin(x)$ , so $v = w^4$ .
First, differentiate $w^4$ with respect to $w$ :
$\frac{d}{dw} (w^4) = 4w^3$

Now, differentiate $w = \sin(x)$ with respect to $x$ :
$\frac{d}{dx} (\sin(x)) = \cos(x)$

Apply the chain rule to combine these derivatives:
$v' = \frac{d}{dx} (\sin^4(x)) = 4 \sin^3(x) \cdot \cos(x)$

Simplify the expression:
$= 6x \cdot \sin^4(x) + 12x^2 \sin^3(x) \cos(x)$

Write the final derivative:
$\frac{d y}{d x} = 6x \sin^4(x) + 12x^2 \sin^3(x) \cos(x)$