Solved on Feb 21, 2024

Find the derivative of $3x(2x+1)^3$ .

STEP 1

Assumptions
1. We need to find the derivative of the function $f(x) = 3x(2x+1)^3$ .
2. We will use the product rule for differentiation, which states that the derivative of a product of two functions is given by: $(fg)' = f'g + fg'$
3. We will also use the chain rule for differentiation, which states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function.

STEP 2

Identify the two functions that make up the product in $f(x)$ . Let $u(x) = 3x$ and $v(x) = (2x+1)^3$ .

STEP 3

Apply the product rule. The derivative of $f(x)$ is: $f'(x) = u'(x)v(x) + u(x)v'(x)$

STEP 4

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

STEP 5

Calculate $u'(x)$ . $u'(x) = 3$

STEP 6

Now, find the derivative of $v(x)$ with respect to $x$ using the chain rule. Let $w(x) = 2x+1$ and $v(x) = w(x)^3$ .

STEP 7

Apply the chain rule to find $v'(x)$ . $v'(x) = \frac{d}{dx}(w(x)^3) \cdot \frac{d}{dx}(2x+1)$

STEP 8

Calculate the derivative of the outer function $w(x)^3$ with respect to $w$ . $\frac{d}{dw}(w^3) = 3w^2$

STEP 9

Calculate the derivative of the inner function $2x+1$ with respect to $x$ . $\frac{d}{dx}(2x+1) = 2$

STEP 10

Substitute $w(x)$ back into the derivative of the outer function. $\frac{d}{dw}(w(x)^3) = 3(2x+1)^2$

STEP 11

Combine the derivatives to find $v'(x)$ . $v'(x) = 3(2x+1)^2 \cdot 2$

STEP 12

Simplify $v'(x)$ . $v'(x) = 6(2x+1)^2$

STEP 13

Now we have both $u'(x)$ and $v'(x)$ . Substitute these into the product rule formula. $f'(x) = u'(x)v(x) + u(x)v'(x)$

STEP 14

Substitute the values of $u'(x)$ , $v(x)$ , $u(x)$ , and $v'(x)$ into the formula. $f'(x) = 3(2x+1)^3 + 3x \cdot 6(2x+1)^2$

STEP 15

Simplify the expression by factoring out common terms if possible. $f'(x) = 3(2x+1)^2 \left[(2x+1) + 2x \right]$

STEP 16

Combine like terms inside the brackets. $f'(x) = 3(2x+1)^2(4x+1)$
The derivative of $f(x) = 3x(2x+1)^3$ is $f'(x) = 3(2x+1)^2(4x+1)$ .

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Find the derivative of $3x(2x+1)^3$ .

STEP 1

STEP 2

Identify the two functions that make up the product in $f(x)$ . Let $u(x) = 3x$ and $v(x) = (2x+1)^3$ .

STEP 3

Apply the product rule. The derivative of $f(x)$ is: $f'(x) = u'(x)v(x) + u(x)v'(x)$

STEP 4

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

STEP 5

Calculate $u'(x)$ . $u'(x) = 3$

STEP 6

Now, find the derivative of $v(x)$ with respect to $x$ using the chain rule. Let $w(x) = 2x+1$ and $v(x) = w(x)^3$ .

STEP 7

Apply the chain rule to find $v'(x)$ . $v'(x) = \frac{d}{dx}(w(x)^3) \cdot \frac{d}{dx}(2x+1)$

STEP 8

Calculate the derivative of the outer function $w(x)^3$ with respect to $w$ . $\frac{d}{dw}(w^3) = 3w^2$

STEP 9

Calculate the derivative of the inner function $2x+1$ with respect to $x$ . $\frac{d}{dx}(2x+1) = 2$

STEP 10

Substitute $w(x)$ back into the derivative of the outer function. $\frac{d}{dw}(w(x)^3) = 3(2x+1)^2$

STEP 11

Combine the derivatives to find $v'(x)$ . $v'(x) = 3(2x+1)^2 \cdot 2$

STEP 12

Simplify $v'(x)$ . $v'(x) = 6(2x+1)^2$

STEP 13

Now we have both $u'(x)$ and $v'(x)$ . Substitute these into the product rule formula. $f'(x) = u'(x)v(x) + u(x)v'(x)$

STEP 14

Substitute the values of $u'(x)$ , $v(x)$ , $u(x)$ , and $v'(x)$ into the formula. $f'(x) = 3(2x+1)^3 + 3x \cdot 6(2x+1)^2$

STEP 15

Simplify the expression by factoring out common terms if possible. $f'(x) = 3(2x+1)^2 \left[(2x+1) + 2x \right]$

STEP 16

Combine like terms inside the brackets. $f'(x) = 3(2x+1)^2(4x+1)$
The derivative of $f(x) = 3x(2x+1)^3$ is $f'(x) = 3(2x+1)^2(4x+1)$ .

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Find the derivative of 3x(2x+1)33x(2x+1)^33x(2x+1)3.

STEP 1

STEP 2

Identify the two functions that make up the product in f(x) f(x) f(x). Let u(x)=3x u(x) = 3x u(x)=3x and v(x)=(2x+1)3 v(x) = (2x+1)^3 v(x)=(2x+1)3.

STEP 3

Apply the product rule. The derivative of f(x) f(x) f(x) is: 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 4

Find the derivative of u(x) u(x) u(x) with respect to x x x. u′(x)=ddx(3x) u'(x) = \frac{d}{dx}(3x) u′(x)=dxd​(3x)

STEP 5

Calculate u′(x) u'(x) u′(x). u′(x)=3 u'(x) = 3 u′(x)=3

STEP 6

Now, find the derivative of v(x) v(x) v(x) with respect to x x x using the chain rule. Let w(x)=2x+1 w(x) = 2x+1 w(x)=2x+1 and v(x)=w(x)3 v(x) = w(x)^3 v(x)=w(x)3.

STEP 7

Apply the chain rule to find v′(x) v'(x) v′(x). v′(x)=ddx(w(x)3)⋅ddx(2x+1) v'(x) = \frac{d}{dx}(w(x)^3) \cdot \frac{d}{dx}(2x+1) v′(x)=dxd​(w(x)3)⋅dxd​(2x+1)

STEP 8

Calculate the derivative of the outer function w(x)3 w(x)^3 w(x)3 with respect to w w w. ddw(w3)=3w2 \frac{d}{dw}(w^3) = 3w^2 dwd​(w3)=3w2

STEP 9

Calculate the derivative of the inner function 2x+1 2x+1 2x+1 with respect to x x x. ddx(2x+1)=2 \frac{d}{dx}(2x+1) = 2 dxd​(2x+1)=2

STEP 10

Substitute w(x) w(x) w(x) back into the derivative of the outer function. ddw(w(x)3)=3(2x+1)2 \frac{d}{dw}(w(x)^3) = 3(2x+1)^2 dwd​(w(x)3)=3(2x+1)2

STEP 11

Combine the derivatives to find v′(x) v'(x) v′(x). v′(x)=3(2x+1)2⋅2 v'(x) = 3(2x+1)^2 \cdot 2 v′(x)=3(2x+1)2⋅2

STEP 12

Simplify v′(x) v'(x) v′(x). v′(x)=6(2x+1)2 v'(x) = 6(2x+1)^2 v′(x)=6(2x+1)2

STEP 13

Now we have both u′(x) u'(x) u′(x) and v′(x) v'(x) v′(x). Substitute these into the product rule formula. 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 14

Substitute the values of 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 formula. f′(x)=3(2x+1)3+3x⋅6(2x+1)2 f'(x) = 3(2x+1)^3 + 3x \cdot 6(2x+1)^2 f′(x)=3(2x+1)3+3x⋅6(2x+1)2

STEP 15

Simplify the expression by factoring out common terms if possible. f′(x)=3(2x+1)2[(2x+1)+2x] f'(x) = 3(2x+1)^2 \left[(2x+1) + 2x \right] f′(x)=3(2x+1)2[(2x+1)+2x]

STEP 16

Combine like terms inside the brackets. f′(x)=3(2x+1)2(4x+1) f'(x) = 3(2x+1)^2(4x+1) f′(x)=3(2x+1)2(4x+1)The derivative of f(x)=3x(2x+1)3 f(x) = 3x(2x+1)^3 f(x)=3x(2x+1)3 is f′(x)=3(2x+1)2(4x+1) f'(x) = 3(2x+1)^2(4x+1) f′(x)=3(2x+1)2(4x+1).

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Find the derivative of $3x(2x+1)^3$ .

Identify the two functions that make up the product in $f(x)$ . Let $u(x) = 3x$ and $v(x) = (2x+1)^3$ .

Apply the product rule. The derivative of $f(x)$ is: $f'(x) = u'(x)v(x) + u(x)v'(x)$

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

Calculate $u'(x)$ . $u'(x) = 3$

Now, find the derivative of $v(x)$ with respect to $x$ using the chain rule. Let $w(x) = 2x+1$ and $v(x) = w(x)^3$ .

Apply the chain rule to find $v'(x)$ . $v'(x) = \frac{d}{dx}(w(x)^3) \cdot \frac{d}{dx}(2x+1)$

Calculate the derivative of the outer function $w(x)^3$ with respect to $w$ . $\frac{d}{dw}(w^3) = 3w^2$

Calculate the derivative of the inner function $2x+1$ with respect to $x$ . $\frac{d}{dx}(2x+1) = 2$

Substitute $w(x)$ back into the derivative of the outer function. $\frac{d}{dw}(w(x)^3) = 3(2x+1)^2$

Combine the derivatives to find $v'(x)$ . $v'(x) = 3(2x+1)^2 \cdot 2$

Simplify $v'(x)$ . $v'(x) = 6(2x+1)^2$

Now we have both $u'(x)$ and $v'(x)$ . Substitute these into the product rule formula. $f'(x) = u'(x)v(x) + u(x)v'(x)$

Substitute the values of $u'(x)$ , $v(x)$ , $u(x)$ , and $v'(x)$ into the formula. $f'(x) = 3(2x+1)^3 + 3x \cdot 6(2x+1)^2$

Simplify the expression by factoring out common terms if possible. $f'(x) = 3(2x+1)^2 \left[(2x+1) + 2x \right]$

Combine like terms inside the brackets. $f'(x) = 3(2x+1)^2(4x+1)$
The derivative of $f(x) = 3x(2x+1)^3$ is $f'(x) = 3(2x+1)^2(4x+1)$ .