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the derivative of something squared is two times something, times the deriavtive of something
LOL i know how to do it i just cant do the whole let u= ...
or you can multiply out an get \[3(x^2+4x+4)=3x^2+12x+12\]\]
i know i multiply by the power take one off then multiply by derivative of the inside of bracket but i need the substitution method
use chain rule
then ignore it, it is a crutch. if you know the chain rule it is entirely unnecessary to say \(u=x+2\) \(y=u^2\) etc
yes, but I believe those were the given instructions
i really would love the substitution method explained
i guess you would say \[u=x+2\] \[y=3u^2\] etc etc, but it is silly
ahhhh okay, then dy/du=6u then du/dx=1 mutliply then substitute u thank you <3
you have a composition of functions 1) first add 2 2) then square the result, multiply by 3 inner function can be written as \(u=x+2\) outer as \(y=3u^2\)
so i substitute u= then y is equal to the new equation with u substituted thank you!
yes, that is correct you get \[6u=6(x+2)\]
do you know the chain rule?
Take the derivative of the inside, take the derivative of the outside leave the inside alone
i know that wy i was struggling with the substitution method i know how to do it straight away though
technically the sub method is the chain rule..