Chain rule - double check me

[Becky4paws]

I'm not sure what to do with the '2' in front of the whole expression.

f(x) = 2(3x+1)^4 (5x-3)^2

2(3x+1)^4 [2(5x-3)(5)] + [8(3x-1)^3(3)](5x-3)^2

2(3x+1)^4 [10(5x-3)] + [24(3x+1)^3](5x-3)^2

2(3x+1)^4(5x-3)[34(3x+1)^3(5x-3)]

 

[galactus]

Product and chain rule:

\(\displaystyle 2(3x+1)^{4}(5x-3)^{2}\)

\(\displaystyle 2(3x+1)^{4}(2)(5x-3)(5)+(5x-3)^{2}(8)(3x+1)^{3}(3)\)

\(\displaystyle 20(3x+1)^{4}(5x-3)^{3}+(5x-3)^{2}(24)(3x+1)\)

Now, factor, rearrange, do whatever you want to get it into a form that suits you.

 

[soroban]

Hello,Becky!

I'm not sure what to do with the '2' in front of the whole expression.

\(\displaystyle \;\;\;f(x) \;= \;2(3x\,+\,1)^4 (5x\,-\,3)^2\;\;\) . . . really?

You may kick yourself . . .

What do you do the the "2" in something like: \(\displaystyle \,y \:=\:2x^3\)
That's right: you bring it along . . . \(\displaystyle y\:=\:2(3x^2)\:=\:6x^2\)


With this problem you have a choice.


Consider it to be: \(\displaystyle \,f(x)\;=\;2\,\cdot\,\left[(3x\,+\,1)^4(5x\,-\,3)^2\right]\)
. . . . . . . . . . . . . . . . . . . .\(\displaystyle 2\;\times\;\text{product}\)

. . . . . . . . . . . . . \(\displaystyle f'(x) \;= \;2\,\cdot\,\left[(3x\.+\,1)^4\cdot2(5x\,-\,3)\cdot5\,+\,(5x\,-\,3)^2\cdot4(3x\,+\,1)^3\cdot3\right]\)


Or consider it to be: \(\displaystyle \,f(x)\;=\;\;\;\;\;\;\;\underbrace{\left[2(3x\,+\,1)^4\right]}\;\;\;\cdot\;\;\;\underbrace{(5x\,-\,3)^2}\)

. . . . . . . . . . . . . . . \(\displaystyle f'(x)\;=\:2(3x\,+\,1)^4\cdot2(5x\,-\,3)\cdot5 \:+\5x\,-\,3)^2\cdot2\cdot4(3x\,+\,1)^3\cdot3\)


They both come out to: \(\displaystyle \,f'(x)\;=\;20(3x\,+\,1)^4(5x\,-\,3)\,+\,24(3x\,+\,1)^3(5x\,-\,3)^2\)


Factor:\(\displaystyle \;f'(x)\;=\;4(3x\,+\,1)^3(5x\,-\,3)\,\cdot\,\left[5(3x\,+\,1)\,+\,6(5x\,-\,3)\right]\)

. . . . . . \(\displaystyle f'(x)\;=\:4(3x\,+\,1)^3(5x\,-\,3)(45x\,-\,13)\)