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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... read more

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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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... read more

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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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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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... read more

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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