Chain rule - double check me

Becky4paws

Junior Member
Joined
Feb 15, 2006
Messages
63
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)]
 
Product and chain rule:

2(3x+1)4(5x3)2\displaystyle 2(3x+1)^{4}(5x-3)^{2}

2(3x+1)4(2)(5x3)(5)+(5x3)2(8)(3x+1)3(3)\displaystyle 2(3x+1)^{4}(2)(5x-3)(5)+(5x-3)^{2}(8)(3x+1)^{3}(3)

20(3x+1)4(5x3)3+(5x3)2(24)(3x+1)\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.
 
Hello,Becky!

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

      f(x)  =  2(3x+1)4(5x3)2    \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: y=2x3\displaystyle \,y \:=\:2x^3
That's right: you bring it along . . . y=2(3x2)=6x2\displaystyle y\:=\:2(3x^2)\:=\:6x^2


With this problem you have a choice.


Consider it to be: f(x)  =  2[(3x+1)4(5x3)2]\displaystyle \,f(x)\;=\;2\,\cdot\,\left[(3x\,+\,1)^4(5x\,-\,3)^2\right]
. . . . . . . . . . . . . . . . . . . .2  ×  product\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: f(x)  =              [2(3x+1)4]            (5x3)2\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: f(x)  =  20(3x+1)4(5x3)+24(3x+1)3(5x3)2\displaystyle \,f'(x)\;=\;20(3x\,+\,1)^4(5x\,-\,3)\,+\,24(3x\,+\,1)^3(5x\,-\,3)^2


Factor:  f(x)  =  4(3x+1)3(5x3)[5(3x+1)+6(5x3)]\displaystyle \;f'(x)\;=\;4(3x\,+\,1)^3(5x\,-\,3)\,\cdot\,\left[5(3x\,+\,1)\,+\,6(5x\,-\,3)\right]

. . . . . . f(x)  =4(3x+1)3(5x3)(45x13)\displaystyle f'(x)\;=\:4(3x\,+\,1)^3(5x\,-\,3)(45x\,-\,13)

 
Top