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Class 12 Tuition > differination of |x| in conditions
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differination of |x| in conditions

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Home Tuitor for Class V to X,Computer,Speaking English,CBSE, ICSE & State Board

Let y = |x| Step 1) Applying first principle ... NOTE:- We shall apply the limit later & now here dx represents delta x. dy/dx = /dx as mentioned we shall apply limit later so currently dy/dx represents Step 2) Multiplying numerator and denominator by we get, /{dx*} Step 3) simplifying...
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Let y=|x| Step 1) Applying first principle ... NOTE:- We shall apply the limit later & now here dx represents delta x. dy/dx = [|x+dx| -|x|]/dx as mentioned we shall apply limit later so currently dy/dx represents [delta y/ delta x ] Step 2) Multiplying numerator and denominator by [|x+dx| + |x|] we get, [(x+dx)^2-x^2]/{dx*[|x+dx| +|x|]} Step 3) simplifying [2x*dx+(dx)^2]/{dx*[|x+dx| +|x|]} As we have not yet applied the limit we cancel a common dx term from numerator and denominator. Step 4) What remains is [2x+dx]/[|x+dx| +|x|] Step 5) Now we apply the limit of dx tends to zero Therefore, dy/dx = x/|x| Now for x>0, dy/dx = 1 for x<0, dy/dx = -1 at x=0 the function is not differentiable as dy/dx is not defined. read less
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Let y=|x|. Step 1 Applying first principle ... NOTE:- We shall apply the limit later. & Currently dx represents delta x. dy/dx=/dx as mentioned we shall apply limit later so currently dy/dx represents Step 2 multiplying numerator and denominator by we get, /{dx*} Step...
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Let y=|x|. Step 1 Applying first principle ... NOTE:- We shall apply the limit later. & Currently dx represents delta x. dy/dx=[|x+dx| -|x|]/dx as mentioned we shall apply limit later so currently dy/dx represents [delta y/ delta x ] Step 2 multiplying numerator and denominator by [|x+dx| + |x|] we get, [(x+dx)^2-x^2]/{dx*[|x+dx| +|x|]} Step 3 simplifying [2x*dx+(dx)^2]/{dx*[|x+dx| +|x|]} as we have not yet applied the limit we can cancel a common dx term from numerator and denominator. Step 4 What remains is [2x+dx]/[|x+dx| +|x|] Step 5 Now we apply the limit of dx tends to zero Therefore, dy/dx=x/|x| Now for x>0, dy/dx=1 for x<0, dy/dx=-1 at x=0 the function is not differentiable as dy/dx is not defined. read less
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"Decoding the World of Physics and Math: 12 Years of Expertise, Powered by a Teaching Enthusiasts"

when x=0- you will get differentiation -1 and when x=0+ you will get differentiation +1 this function is not differentiable at x=0
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Mathematics Professor

Since the absolute value is defined by cases, |x|={x?xif x?0;if x0, x0, for ?x sufficiently close to 0 we will have x+?x>0. So f(x)=|x|=x, and f(x+?x)=|x+?x|=x+?x; plugging that into the limit, we have: lim?x?0f(x+?x)?f(x)?x=lim?x?0|x+?x|?|x|?x=lim?x?0(x+?x)?x?x. You should be able to finish it now. For...
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Since the absolute value is defined by cases, |x|={x?xif x?0;if x<0, it makes sense to deal separately with the cases of x>0, x<0, and x=0. For x>0, for ?x sufficiently close to 0 we will have x+?x>0. So f(x)=|x|=x, and f(x+?x)=|x+?x|=x+?x; plugging that into the limit, we have: lim?x?0f(x+?x)?f(x)?x=lim?x?0|x+?x|?|x|?x=lim?x?0(x+?x)?x?x. You should be able to finish it now. For x<0, for ?x sufficiently close to zero we will have x+?x<0; so f(x)=?x and f(x+?x)=?(x+?x). It should again be easy to finish it. The tricky one is x=0. I suggest using one-sided limits. For the limit as ?x?0+, x+?x=?x>0; for ?x?0?, x+?x=?x<0; the (one-sided) limits should now be straightforward. good luck read less
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IIT/BITSAT Decoded !!

Differentiation of |x| is |x|/x , so for positive & negative x you can easily put x with appropriate sign to get the result.
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Tutor

if x>0 it will be 1 and if x<0 then x=-1
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Take positive and negative sign and than differentiate and is 1 and -1 if x>0and x<0.
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Trainer

(d/dx) IxI = IxI/x, which is not continuous.
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If x>=0, Then d|x|/d|y|=1 or else -1
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All that glitter is not gold

1 or -1
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