anonymous
  • anonymous
hi all, I m taking this course to pursue a career in physics. 1) How to know whether a given function is differentiable ? 2) Also, as far as i know, calculus is an extension of geometry and algebra to deal with rate of change and area under curves. But I see some examples of certain functions (like f(x) = |x|, whose graph is a straight line) being differentiated. Could some one help me with this ?
OCW Scholar - Single Variable Calculus
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
You only know that a function is differentiable by taking the derivate of the function. But formally the following must be true for function f(x) to be differentiable at x0:\[\lim_{x \rightarrow x _{0}}f(x)-f(x_{0})=0\]eg. the function f must be continous at x0. Then\[\lim_{x \rightarrow x _{0}^{+}}\frac{ f(x)-f(x_{0}) }{ x-x_{0} }=\lim_{x \rightarrow x_{0}^{-}}\frac{ f(x)-f(x_{0}) }{ x-x_{0} }=\lim_{x \rightarrow x_{0}}\frac{ f(x)-f(x_{0}) }{ x-x_{0} }\]Which means that the right derivative at x0 must be the same as the left derivative at x0 which is the derivative at x0. Couple of examples that are not diffrentiable everywhere: |x| is not diffrentiable at 0 even though it's continous there because it's right derivative at 0 is 1 and it's left derivative at 0 is -1. However |x| is differentable at R/{0} and so one can calculate it's derivate as:\[\frac{ d }{ dx }\left| x \right| = 1, x>0\]and\[\frac{ d }{ dx }\left| x \right|= -1, x<0\]So the derivate itself is not continous. The function that is 0 everywhere except at 0 where it's 1 has both left and right derivatives at 0 and the value of those derivatives is 0, but it has no derivative at 0 nor is it continous at 0. Hope this helps.
anonymous
  • anonymous
Thanks, it helped a lot.
anonymous
  • anonymous
I thought about it for a while. To sum up (correct me if i am wrong), a function is differentiable at a point, if it has a well defined tangent at that point. If the slope of the function f(x) = |x|, at x = 0 had been a unique value, it would have been a fully differentiable function.

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anonymous
  • anonymous
Yes that's true. The derivate of a function gives the slope of the tangent at a given point. If the tangent is unambigous in the given point then the derivate also exist in that point..
anonymous
  • anonymous
I'd like to further add that if the tangent is vertical then the derivative is infinite but might still exist. For example the equation\[x^{2}+y ^{2}=a ^{2}\]describes an origo centered circle with radius of a and it has a derivate dy/dx that at the points (-a, 0) and (0, a) goes to infinity, but at those point dx/dy is 0 so the tangent is well defined as x=-a and x=a respectively.

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