Chain Rule

May 2010
3
0
I'm doing a chain rule and i am having trouble factoring it out, help would be appreciated.
A step by step explanation would help
Heres what i got so far;

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

I can't seem to factor it out, and i should know how =/
 

skeeter

MHF Helper
Jun 2008
16,216
6,764
North Texas
I'm doing a chain rule and i am having trouble factoring it out, help would be appreciated.
A step by step explanation would help
Heres what i got so far;

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

I can't seem to factor it out, and i should know how =/
\(\displaystyle -6(3-2x)^2(5x+2)^4 + 20(3-2x)^3(5x+2)^3\)

common factors of both terms are \(\displaystyle 2(3-2x)^2(5x+2)^3\) ...

\(\displaystyle 2(3-2x)^2(5x+2)^3[-3(5x+2) + 10(3-2x)]\)

distribute and combine terms inside the [brackets] to finish
 
May 2010
3
0
Thankyou, also I have (4)(x^2+2)^2-(4x)2(x^2+2)(2x) / (x^2+2)^4 could you help me set that to 0 i think i get

4(x^2+2) [(x^2+2)+4x^2(4x)] / (x^2+2)^4 right?

Then i gotta set it all to 0
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
Thankyou, also I have (4)(x^2+2)^2-(4x)2(x^2+2)(2x) / (x^2+2)^4 could you help me set that to 0 i think i get

4(x^2+2) [(x^2+2)+4x^2(4x)] / (x^2+2)^4 right?
Almost- you dropped a "-" and left a "4x". It should be 4(x^2+ 2)[(x^2+2)- 4x^2]/(x^2+ 2)^4

Then i gotta set it all to 0[/QUOTE]
Once you have set it equal to 0, you can immediately multiply through by that (4x^2+ 2)^4 in the denominator and get rid of it- a fraction is equal to 0 if and only if the numerator is equal to 0.
 
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