Calculus: chain rule with product/quotient rule (step by step please?)

[Silencher]

Hey all. Below are the various problems I did and how I did them. Can someone go through what I did and show me where I messed up in which step and how I messed up? I really don't understand what I'm doing wrong.

(1.) WRONG


derive: y = (3x-5)^4 * (2-x^4)^4


Step 1: Find g/h


g(x): (3x-5)^4
h(x): (2-x^4)^4
g'(x): (12)(3x-5)^3
h'(x): (16x^3)(2-x^4)^3


((3x-5)^4) * (16x^3)((2-x^4)^3) + ((2-x^4)^4)(12)((3x-5)^3)<--my answer


4(3x-5)^3 (2-x^4)^3(6+20x^3 -15x^4)<--correct answer


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(2.) WRONG


Derive: y = x(x^3+9)^4


(4x(x^3+9)^3)(3x^2)<--my answer


(x^3+9)^3(13x^3+9) <---right answer


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(3.) WRONG


Derive: y = (3x-3)^5 * (5-x^4)^3


(3)(5)((3x-3)^4)((5-x^4)^3) +((3x-3)^5)(4x^3)(3(5-x^4)^2)


(15)((3x-3)^4)((5-x^4)^3)+((3x-3)^5)(12x^3)(5-x^4)^2)<--myanswer(simplified)


(243)(x-1)^4(5-x^4)^2(25+12x^3-17x^4)<---correct answer
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(7.) WRONG


y = ((4x+1)^4) / ((3x+1)^3)


Step 1: Set g/h's


g(x): (4x+1)^4
h(x): (3x+1)^3
g'(x): 16((4x+1)^3) <--simplifiedfrom: (4)(4)((4x+1)^3)
h'(x): 9(3x+1)^2)<--simplified from3)(3)((3x+1)^2)
(h(x))^2: ((3x+1)^3)^2)


Step 2: Input into quotient formula:


(16)((4x+1)^3)((3x+1)^3) -(9)((3x+1)^2) ((4x+1)^4)
____________________________________
(((3x+1)^3)^2)


Step 3: Simplify/my answer:


((16)((12x^2+7x+1)^3)) -((9)((3x+1)^2)((4x+2)^4)
____________________________________
(((3x+1)^3)^2)


=below is the correct answer=


(4x+1)^3 * (12x+7)
________________
(3x+1)^4

 

[lookagain]

I will show work/comments on just the first one.

(1.) WRONG


derive: y = (3x-5)^4 * (2-x^4)^4


Step 1: Find g/h $\displaystyle \ \ \ \$ No, you mean g*h.


g(x): (3x - 5)^4


h(x): (2 - x^4)^4


g'(x): (12)(3x - 5)^3


h'(x): (-16x^3)(2 - x^4)^3 $\displaystyle \ \ \ \$ You were missing a negative sign for the coefficient on x^3. (I typed one in here, but not in the next line.)


((3x - 5)^4) * (16x^3)((2 - x^4)^3) + ((2 - x^4)^4)(12)((3x - 5)^3) <--my answer $\displaystyle \ \ \$ The answer took it further and factored out the common binomial factors.


4(3x - 5)^3 (2 - x^4)^3(6 + 20x^3 - 15x^4) <--correct answer

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$\displaystyle (3x - 5)^4 (-16x^3)(2 - x^4)^3 \ + \ (2 - x^4)^4(12)(3x - 5)^3 \ =$

$\displaystyle 4(3x - 5)^3 (2 - x^4)^3[-4x^3(3x - 5) \ + \ 3(2 - x^4)] \ =$


$\displaystyle 4(3x - 5)^3(2 - x^4)^3(-12x^4 + 20x^3 + 6 - 3x^4) \ =$

$\displaystyle 4(3x - 5)^3(2 - x^4)^3(6 + 20x^3 - 15x^4)$

 

[JeffM]

Hey all. Below are the various problems I did and how I did them. Can someone go through what I did and show me where I messed up in which step and how I messed up? I really don't understand what I'm doing wrong.

(2.) WRONG


Derive: y = x(x^3+9)^4


(4x(x^3+9)^3)(3x^2)<--my answer


(x^3+9)^3(13x^3+9) <---right answer

The way to avoid mistakes is to use substitutions and the chain rule

$\displaystyle Find\ y'\ given\ y = x(x^3+9)^4.$

$\displaystyle u = x,\ v = x^3 + 9,\ and\ w = v^4 \implies y = uw \implies \dfrac{dy}{dx} = \dfrac{du}{dx} * w + u * \dfrac{dw}{dx}.$ Easy, basic product rule.

$\displaystyle u = x \implies \dfrac{du}{dx} = 1.$

$\displaystyle v = x^3 + 9 \implies \dfrac{dv}{dx} = 3x^2.$ Easy, power rule.

$\displaystyle w = v^4 \implies \dfrac{dw}{dv} = 4v^3 \implies \dfrac{dw}{dx} = \dfrac{dw}{dv} * \dfrac{dv}{dx}.$ Easy, chain rule.

So $\displaystyle \dfrac{dy}{dx} = 1 * v^4 + u * 4v^3 * 3x^2 = v^3(v + 12ux^2) = (x^3 + 9)(x^3 + 9 + 12x * x^2) = (x^3 + 9)(13x^3 + 9).$

Now these substitutions abd back substitutions take more time, but they are easy and make it harder to make mistakes.

 

[soroban]

Hello, Silencher!

$\displaystyle (7)\;y \:=\: \dfrac{(4x+1)^4}{(3x+1)^3}$

$\displaystyle \text{We have: }\;y \:=\: \dfrac{\overbrace{(4x+1)^4}^{f}}{\underbrace{(3x+1)^3}_g}$

$\displaystyle \text{Then: }\;\;y' \;=\;\dfrac{\overbrace{(3x+1)^3}^g\cdot\overbrace{4(4x+1)^3\cdot 4}^{f'} - \overbrace{(4x+1)^4}^f\cdot\overbrace{3(3x+1)^2 \cdot 3}^{g'}}{\underbrace{\left[(3x+1)^3\right]^2}_{g^2}}$

$\displaystyle \text{Factor: }\:y' \;=\;\dfrac{(3x+1)^2(4x+1)^3\cdot\big[16(3x+1) - 9(4x+1)\big]}{(3x+1)^6}$

. . . . . . $\displaystyle y' \;=\;\dfrac{(4x+1)^3\cdot\big[48x + 16 - 36x - 9\big]}{(3x+1)^4}$

. . . . . . $\displaystyle y' \;=\;\dfrac{(4x+1)^3(12x+7)}{(3x+1)^4}$