# how to prove a function is differentiable at a point

Greatest Integer Function [x] Going by same Concept Ex 5.2, 10 Prove that the greatest integer function defined by f (x) = [x], 0 < x < 3 is not differentiable at =1 and = 2. For $$t \neq 0$$, we have $\frac{h(t \cos \theta, t \sin \theta) – h(0,0)}{t}= \frac{ \cos^2 \theta \sin \theta}{\cos^6 \theta + \sin^2 \theta}$ which is constant as a function of $$t$$, hence has a limit as $$h \to 0$$. Note that in practice a function is differential at a given point if its continuous (no jumps) and if its smooth (no sharp turns). Sal analyzes a piecewise function to see if it's differentiable or continuous at the edge point. &= \lim_{h \to 0}h \sin (1/|h|) =0. Another point of note is that if f is differentiable at c, then f is continuous at c. if and only if f' (x 0 -) = f' (x 0 +) . the absolute value for $$\mathbb R$$. Such ideas are seen in university mathematics. A function having partial derivatives which is not differentiable. 0 & \text{ if }(x,y) = (0,0).\end{cases}\] $$f$$ is obviously continuous on $$\mathbb R^2 \setminus \{(0,0)\}$$. \frac{\partial f}{\partial x_i}(\mathbf{a}) &= \lim\limits_{h \to 0} \frac{f(\mathbf{a}+h \mathbf{e_i})- f(\mathbf{a})}{h}\\ $$f$$ is also continuous at $$(0,0)$$ as for $$(x,y) \neq (0,0)$$ $\left\vert (x^2+y^2)\sin\left(\frac{1}{\sqrt{x^2+y^2}}\right) \right\vert \le x^2+y^2 = \Vert (x,y) \Vert^2 \mathrel{\mathop{\to}_{(x,y) \to (0,0)}} 0$ $$f$$ is also differentiable at all $$(x,y) \neq (0,0)$$. A function f is differentiable at a point c if. Say, if the function is convex, we may touch its graph by a Euclidean disc (lying in the épigraphe), and in the point of touch there exists a derivative. Both of these derivatives oscillate wildly near the origin. If limits from the left and right of that point are the same it's diferentiable. Then solve the differential at the given point. \left(\frac{1}{\sqrt{x^2+y^2}}\right)}{\sqrt{x^2+y^2}}. The partial maps $$x \mapsto g(x,0)$$ and $$y \mapsto g(0,y)$$ are always vanishing. However, $$h$$ is not differentiable at the origin. In this video I go over the theorem: If a function is differentiable then it is also continuous. Example of a Nowhere Differentiable Function \frac{\partial f}{\partial x}(x,y) &= 2 x \sin Then $$f$$ is continuously differentiable if and only if the partial derivative functions $$\frac{\partial f}{\partial x}(x,y)$$ and $$\frac{\partial f}{\partial y}(x,y)$$ exist and are continuous. exists. A function is said to be differentiable if the derivative exists at each point in its domain. 'http':'https';if(!d.getElementById(id)){js=d.createElement(s);js.id=id;js.src=p+'://platform.twitter.com/widgets.js';fjs.parentNode.insertBefore(js,fjs);}}(document, 'script', 'twitter-wjs'); Is it okay to just show at the point of transfer between the two pieces of the function that f(x)=g(x) and f'(x)=g'(x) or do I need to show limits and such. Differentiability at a point: algebraic (function is differentiable) Differentiability at a point: algebraic (function isn't differentiable) Practice: Differentiability at a point: ... And we talk about that in other videos. Definition 1 We say that a function $$f : \mathbb R^2 \to \mathbb R$$ is differentiable at $$\mathbf{a} \in \mathbb R^2$$ if it exists a (continuous) linear map $$\nabla f(\mathbf{a}) : \mathbb R^2 \to \mathbb R$$ with $\lim\limits_{\mathbf{h} \to 0} \frac{f(\mathbf{a}+\mathbf{h})-f(\mathbf{a})-\nabla f(\mathbf{a}).\mathbf{h}}{\Vert \mathbf{h} \Vert} = 0$. Watch Queue Queue. Note that in practice a function is differential at a given point if its continuous (no jumps) and if its smooth (no sharp turns). Let’s have a look to the directional derivatives at the origin. \end{align*} So, a function is differentiable if its derivative exists for every $$x$$-value in its domain. We now consider the converse case and look at $$g$$ defined by So f is not differentiable at x = 0. Would you like to be the contributor for the 100th ring on the Database of Ring Theory? the question is too vague to be able to give a meaningful answer. And then right when x is equal to one and the value of our function is zero it looks something like this, it looks something like this. Differentiable Function: A function is said to be differentiable at a point if and only if the derivative of the given function is defined at that point. Therefore, $$h$$ has directional derivatives along all directions at the origin. Finally $$f$$ is not differentiable. In other words: The function f is diﬀerentiable at x if lim h→0 f(x+h)−f(x) h exists. !function(d,s,id){var js,fjs=d.getElementsByTagName(s)[0],p=/^http:/.test(d.location)? In the same way, one can show that $$\frac{\partial f}{\partial y}$$ is discontinuous at the origin. f(x)=[x] is not continuous at x = 1, so it’s not differentiable at x = 1 (there’s a theorem about this). Hence $$h$$ is continuously differentiable for $$(x,y) \neq (0,0)$$. Want to be posted of new counterexamples? 0 & \text{ if }(x,y) = (0,0)\end{cases}\] has directional derivatives along all directions at the origin, but is not differentiable at the origin. To be able to tell the differentiability of a function using graphs, you need to check what kind of shape the function takes at that certain point.If it has a smooth surface, it implies it’s continuous and differentiable. If it is a direct turn with a sharp angle, then it’s not continuous. We recall some definitions and theorems about differentiability of functions of several real variables. A function f is not differentiable at a point x0 belonging to the domain of f if one of the following situations holds: (i) f has a vertical tangent at x 0. Definition 3 Let $$f : \mathbb R^n \to \mathbb R$$ be a real-valued function. Away from the origin, one can use the standard differentiation formulas to calculate that Then solve the differential at the given point. As we head towards x = 0 the function moves up and down faster and faster, so we cannot find a value it is "heading towards". Thus, the graph of f has a non-vertical tangent line at (x,f(x)). Continuity of the derivative is absolutely required! Press J to jump to the feed. Continuity of the derivative is absolutely required! $g(x,y)=\begin{cases}\frac{xy}{\sqrt{x^2+y^2}} & \text{ if } (x,y) \ne (0,0)\\ 0 & \text{ if }(x,y) = (0,0).\end{cases}$ For all $$(x,y) \in \mathbb R^2$$ we have $$x^2 \le x^2+y^2$$ hence $$\vert x \vert \le \sqrt{x^2+y^2}=\Vert (x,y) \Vert$$. Hence $$g$$ has partial derivatives equal to zero at the origin. New comments cannot be posted and votes cannot be cast. \frac{f(h,0)-f(0,0)}{h}\\ &= \lim_{h \to 0}\frac{h^2 \sin (1/|h|)-0}{h} \\ For example, the derivative with respect to $$x$$ along the $$x$$-axis is $$\frac{\partial f}{\partial x}(x,0) = 2 x \sin Consider the function defined on \(\mathbb R^2$$ by Continue Reading. If you get two numbers, infinity, or other undefined nonsense, the function is not differentiable. Differentiate it. If a function is continuous at a point, then is differentiable at that point. 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