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Class 12 Tuition > Learn Class XI-XII Tuition (PUC) > differination of |x| in conditions
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differination of |x| in conditions

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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 = /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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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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"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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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...
read more
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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if x>0 it will be 1 and if x<0 then it will be -1 and at x=0 derivative is not defined.
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i am a tution teacher

In differination ill give signum function 1when X is positive and -1 when it is negative
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Teaching is my passion!!!!

if x>0 it will be 1 and if x<0 then it will be -1 and at x=0 derivative is not defined
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Trainer

(d/dx) IxI = IxI/x, which is not continuous.
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Conditions under which one may differentiate term by term a convergent sequence of functions are to be found in the literature.f Some of these conditions are obtained as corollaries of theorems about term wise integration, and consequently contain in the hypothesis the assumption that the derived...
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Conditions under which one may differentiate term by term a convergent sequence of functions are to be found in the literature.f Some of these conditions are obtained as corollaries of theorems about term wise integration, and consequently contain in the hypothesis the assumption that the derived sequence converges. Without this assumption, sufficient conditions for term wise differentiation may be obtained from the fundamental theorem on reversing the order of iterated limits. read less
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