Multiple derivative rules?

[confused_07]

Do you use multiple rules to find the derivative?
Ex: \(\displaystyle f[x] = (x-2)^2 (x+3)^2\)

I've got:

Product Rule-
\(\displaystyle f'[x] = (x^2-4x+2)(2x+6) + (2x-4)(x^2+6x+9)
= (2x^3-2x^2-20x+12) + (2x^3+8x^2-6x-36)
= 4x^3+6x^2-26x-24\)

Would I use the Power Rule next, or is it done?

 

[arthur ohlsten]

f[x]= [x-2]^2[x+3]^2
let w=[x-2]^2
then dw/dx= 2[x-2]

let z=[x+3]^2
then dz/dx= 2[x+3]

f[x]=wz take the derivative
df/dx = w dz/dx + z dw/dx
df/dx= 2[x-2]^2[x+3] +[x+3]^2[2][x-2]
df/dx= 2[x-2][x+3][ [x-2]+[x+3] ]
df/dx= 2[x-2][x+3][2x+1] answer
or by multiplying
df/dx=2[x^2+x-6][2x+1]
df/dx=2[2x^3+2x^2-12x+x^2+x-6]
df/dx=2[2x^3+3x^2-11x-6]
df/dx=4x^3+6x^2-22x-12 answer

Arthur
please check for errors

 

[soroban]

Hello, confused_07!

If you're going to multiply-out parts of the function,
. . why not multiply it all out?

\(\displaystyle f(x)\:=\:x^4\,+\,2x^3\,-\,11x^2\,-\,12x\,+\,36\)
. . Then you don't have to worry about "multiple rules" . . .

Differentiate: \(\displaystyle \:f(x)\:=\x\,-\,2)^2(x\,+\,3)^2\)

Apply both the Product Rule and the Chain Rule at the same time.

\(\displaystyle f'(x) \;=\;(x\,-\,2)^2\cdot2(x\,+\,3) \,+\, 2(x\,-\,2)\cdot(x\,+\,3)^2\)

Factor: \(\displaystyle \:2(x\,-\,2)(x\,+\,3)\,[(x\,-\,2)\,+\,(x\,+\,3)]\)


Answer: \(\displaystyle \:f'(x) \;= \;\fbox{2(x\,-\,2)(x\,+\,3)(2x\,+\,1)}\)

 

[arthur ohlsten]

I assume you are studying the product rule.
If you are I suggest you use the product rule and then as a check multiply out the function to check on your answer
let w=[x-2]^2
then dw/dx=2[x-2]

let z= [x+3]^2
dz/dx=2[x+3]

y=wz
the derivative is the 1st times the derivative of the second plus the 2nd times the derivative of the first.
dy/dx=w dz/dx + z dw/dx
dy/dx=[x-2]^2 [2][x+3] + [x+3]^2[2][x-2] answer but simplify to
dy/dx = 2[x-2][x+3][x-2 +x+3]
dy/dx = 2[x-2][x+3][2x+1] answer

I suggest you then check the answer by multiplying out the original function and take the resuting equations derivative.