Continuously differentiable vector-valued functions. The function is differentiable from the left and right. This should be rather obvious, but a function that contains a discontinuity is not differentiable at its discontinuity. Then, using Ito's Lemma and integrating both sides from $t_0$ to $t$ reveals that, $$X_t=X_{t_0}e^{(\alpha-\beta^2/2)(t-t_0)+\beta(W_t-W_{t_0})}$$. But a function can be continuous but not differentiable. When this limit exist, it is called derivative of #f# at #a# and denoted #f'(a)# or #(df)/dx (a)#. The first type of discontinuity is asymptotic discontinuities. ? by Lagranges theorem should not it be differentiable and thus continuous rather than only continuous ? That is, the graph of a differentiable function must have a (non-vertical) tangent line at each point in its domain, be relatively "smooth" (but not necessarily mathematically smooth), and cannot contain any breaks, corners, or cusps. Of course, you can have different derivative in different directions, and that does not imply that the function is not differentiable. If a function is differentiable and convex then it is also continuously differentiable. Trump has last shot to snatch away Biden's win, Cardi B threatens 'Peppa Pig' for giving 2-year-old silly idea, These 20 states are raising their minimum wage, 'Super gonorrhea' may increase in wake of COVID-19, ESPN analyst calls out 'young African American' players, Visionary fashion designer Pierre Cardin dies at 98, Cruz reportedly got $35M for donors in last relief bill, More than 180K ceiling fans recalled after blades fly off, Bombing suspect's neighbor shares details of last chat, Biden accuses Trump of slow COVID-19 vaccine rollout. Example 1: Differentiable ⇒ Continuous. But it is not the number being differentiated, it is the function. 2. inverse function. On the other hand, if you have a function that is "absolutely" continuous (there is a particular definition of that elsewhere) then you have a function that is differentiable practically everywhere (or more precisely "almost everywhere"). But there are functions like $\cos(z)$ which is analytic so must be differentiable but is not "flat" so we could again choose to go along a contour along another path and not get a limit, no? A discontinuous function is not differentiable at the discontinuity (removable or not). Note: The converse (or opposite) is FALSE; that is, … an open subset of , where ≥ is an integer, or else; a locally compact topological space, in which k can only be 0,; and let be a topological vector space (TVS).. The next graph you have is a cube root graph shifted up two units. Learn how to determine the differentiability of a function. Anyhow, just a semantics comment, that functions are differentiable. Theorem: If a function f is differentiable at x = a, then it is continuous at x = a. Contrapositive of the above theorem: If function f is not continuous at x = a, then it is not differentiable at x = a. Both those functions are differentiable for all real values of x. well try to see from my perspective its not exactly duplicate since i went through the Lagranges theorem where it says if every point within an interval is continuous and differentiable then it satisfies the conditions of the mean value theorem, note that it defines it for every interval same does the work cauchy's theorem and fermat's theorem that is they can be applied only to closed intervals so when i faced question for open interval i was forced to ask such a question, https://math.stackexchange.com/questions/1280495/when-is-a-continuous-function-differentiable/1280504#1280504. True. for every x. 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