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calc help!!!! how do you differentiate absolute values?!?!??!?!

Mathematics
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y= abs(3x-1)
  • phi
when x > 1/3 this is the same as y= 3x-1 when x< 1/3 it is y= -3x+1
so what rule can u use for the absolute value problems or can u do it on ur calculator

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Other answers:

bc i know you cant use the product rule or the quotient rule or the power rule it seems like or the chain rule
  • phi
one other thing, dy/dx is undefined at x= 1/3 use the power rule d/dx a x^n = a n x^(n-1)
how would u use the power rule for the abs(3x-1)
would u set (3x-1)^1 ?
  • phi
you get 3 different answers, depending on x x> 1/3 d/dx (y= 3x-1) x= 1/3 undefined x<1/3 d/dx (y= -3x+1) d/dx y = dy/dx d/dx (3x^1 - 1) = d/dx 3x^1 + d/dx (-1) can you finish?
yeah now i can oh ok and also i have another question for absolute values differentiate y= abs(x^2-5x+4) would u jst factor it out first?
and use the product rule?
  • phi
for abs(x^2-5x+4) you have to find the values of x where this stays positive and the regions where it is negative (then replace abs with - )
  • phi
First finish the first problem, then repost this quadratic, because it is more complicated, and might take time to explain
kk
if I may add, one trick to differentiating absolute value can be to use the definition\[|x|=\sqrt{x^2}\]let\[u=x^2\]then\[\frac d{dx}\sqrt u=\frac{u'}{2\sqrt u}=\frac{2x}{2|x|}=\frac x{|x|}\]
can u differentiate it on your calculator?
I rely on my brain for differentiation usually, I've lost track of what handheld calculators are good for besides graphing and handling ugly nmbers
numbers*
i dont think my teacher has taught us how to differentiate absolute values yet but she assigned us an assignment with like 2 absolute value problems im still a bit confused on how to differentiate it
I just showed you one way
what r the other ways
#stillconfused
  • phi
btw, turing I think you should tweak your rule to x dx/|x|
@abannavong, another way to find the derivative is to set up the difference quotient and multiply by the conjugate of the numerator (|x+h|-|x|)/h-->((|x+h|-|x|)(|x+h|+|x|))/(h*(|x+h|+|x|)) Hint: When you square an absolute value, you can drop the absolute value signs
oh ok thanks @L.T.

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