# how to prove a function is differentiable at a point

From the Fig. So it is not differentiable over there. Ex 5.2, 10 (Introduction) Greatest Integer Function f(x) = [x] than or equal to x. 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. Want to be posted of new counterexamples? 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. 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. \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}\\ If f is differentiable at a point x 0, then f must also be continuous at x 0.In particular, any differentiable function must be continuous at every point in its domain. Consequently, $$g$$ is a continuous function. 10.19, further we conclude that the tangent line is vertical at x = 0. Both of these derivatives oscillate wildly near the origin. The partial maps $$x \mapsto g(x,0)$$ and $$y \mapsto g(0,y)$$ are always vanishing. Analyze algebraic functions to determine whether they are continuous and/or differentiable at a given point. To be differentiable at a certain point, the function must first of all be defined there! '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'); Differentiate it. You want to find rings having some properties but not having other properties? $f(x,y)=\begin{cases}(x^2+y^2)\sin\left(\frac{1}{\sqrt{x^2+y^2}}\right) & \text{ if } (x,y) \ne (0,0)\\ If you get a number, the function is differentiable. Continuity of the derivative is absolutely required! 0 & \text{ if }(x,y) = (0,0).\end{cases}$ $$f$$ is obviously continuous on $$\mathbb R^2 \setminus \{(0,0)\}$$. Free ebook http://tinyurl.com/EngMathYT A simple example of how to determine when a function is differentiable. Theorem 2 Let $$f : \mathbb R^2 \to \mathbb R$$ be differentiable at $$\mathbf{a} \in \mathbb R^2$$. Show that the function is continuous at that point (doesn't have a hole or asymptote or something) and that the limit as x (or whatever variable) approaches that point from all sides is the same as the value of the function at that point. Or subscribe to the RSS feed. A function is said to be differentiable if the derivative exists at each point in its domain. The partial derivatives of $$f$$ are zero at the origin. \left(\frac{1}{\sqrt{x^2+y^2}}\right)}{\sqrt{x^2+y^2}}. Differentiate it. 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 $$h \to 0$$. $$\mathbb R^2$$ and $$\mathbb R$$ are equipped with their respective Euclidean norms denoted by $$\Vert \cdot \Vert$$ and $$\vert \cdot \vert$$, i.e. \frac{\partial f}{\partial x}(0,0) &= \lim_{h \to 0} Nowhere Differentiable. We also have $\frac{\partial g}{\partial x}(x,y) = \frac{y^3}{(x^2+y^2)^{\frac{3}{2}}}, \frac{\partial g}{\partial x}(0,y) = \text{sign}(y)$ which proves that $$\frac{\partial g}{\partial x}$$ is not continuous at the origin avoiding any contradiction with theorem 1. Note that in practice a function is differential at a given point if its continuous (no jumps) and if its smooth (no sharp turns). Post all of your math-learning resources here. We now consider the converse case and look at $$g$$ defined by Continue Reading. Rather, it serves to illustrate how well this method of approximation works, and to reinforce the following concept: exists for every c in (a, b). In the same way, one can show that $$\frac{\partial f}{\partial y}$$ is discontinuous at the origin. 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. Let’s fix $$\mathbf{v} = (\cos \theta, \sin \theta)$$ with $$\theta \in [0, 2\pi)$$. 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". A function having partial derivatives which is not differentiable. Theorem 1 Let $$f : \mathbb R^2 \to \mathbb R$$ be a continuous real-valued function. In fact $$h$$ is not even continuous at the origin as we have $h(x,x^3) = \frac{x^2 x^3}{x^6 + (x^3)^2} = \frac{1}{x}$ for $$x \neq 0$$. Would you like to be the contributor for the 100th ring on the Database of Ring Theory? Let’s have a look to the directional derivatives at the origin. Definition 2 Let $$f : \mathbb R^n \to \mathbb R$$ be a real-valued function. if and only if f' (x 0 -) = f' (x 0 +) . \frac{f(h,0)-f(0,0)}{h}\\ Hence $$h$$ is continuously differentiable for $$(x,y) \neq (0,0)$$. Hence $$g$$ has partial derivatives equal to zero at the origin. We want to show that: lim f(x) − f(x 0) = 0. x→x 0 This is the same as saying that the function is continuous, because to prove that a function was continuous we’d show that lim f(x) = f(x 0). We focus on real functions of two real variables (defined on $$\mathbb R^2$$). Such ideas are seen in university mathematics. Maybe, it allows to prove something about the set of points where there is no derivative, not only that it has Lebesgue measure $0$. Similarly, f is differentiable on an open interval (a, b) if. the question is too vague to be able to give a meaningful answer. If you get two numbers, infinity, or other undefined nonsense, the function is not differentiable. This counterexample proves that theorem 1 cannot be applied to a differentiable function in order to assert the existence of the partial derivatives. Basically, f is differentiable at c if f' (c) is defined, by the above definition. the absolute value for $$\mathbb R$$. Answer to: How to prove that a function is differentiable at a point? Consider the function defined on $$\mathbb R^2$$ by to show that a function is differentiable, show that the limit exists. 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 $\lim\limits_{\mathbf{h} \to 0} \frac{f(\mathbf{a}+\mathbf{h})-f(\mathbf{a})-\nabla f(\mathbf{a}).\mathbf{h}}{\Vert \mathbf{h} \Vert} = 0$. Definition 3 Let $$f : \mathbb R^n \to \mathbb R$$ be a real-valued function. Example Let's have another look at our first example: $$f(x) = x^3 + 3x^2 + 2x$$. \end{align*} For two real variable functions, $$\frac{\partial f}{\partial x}(x,y)$$ and $$\frac{\partial f}{\partial y}(x,y)$$ will denote the partial derivatives. $$f$$ 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 $$(x,y) \neq (0,0)$$. As in the case of the existence of limits of a function at x 0, it follows that. (ii) The graph of f comes to a point at x 0 (either a sharp edge ∨ or a sharp peak ∧ ) (iii) f is discontinuous at x 0. 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. Watch Queue Queue Thus, the graph of f has a non-vertical tangent line at (x,f(x)). How to prove a piecewise function is both continuous and differentiable? : The function is differentiable from the left and right. for $$x \neq 0$$, where $$\text{sign}(x)$$ is $$\pm 1$$ depending on the sign of $$x$$. Then solve the differential at the given point. Now some theorems about differentiability of functions of several variables. Afunctionisdiﬀerentiable at a point if it has a derivative there. 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. A function f is differentiable at a point c if. In other words: The function f is diﬀerentiable at x if lim h→0 f(x+h)−f(x) h exists. &= \lim_{h \to 0}h \sin (1/|h|) =0. \frac{\partial f}{\partial y}(x,y) &= 2 y \sin We recall some definitions and theorems about differentiability of functions of several real variables. \end{align*} Do you know the definition of derivate by limit? Therefore, $$h$$ has directional derivatives along all directions at the origin. Similarly, $$\vert y \vert \le \Vert (x,y) \Vert$$ and therefore $$\vert g(x,y) \vert \le \Vert (x,y) \Vert$$. This article provides counterexamples about differentiability of functions of several real variables. If a function is continuous at a point, then is differentiable at that point. Example of a Nowhere Differentiable Function f(x)=[x] is not continuous at x = 1, so it’s not differentiable at x = 1 (there’s a theorem about this). 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So f is not differentiable at x = 0. This last inequality being also valid at the origin. 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. In this video I go over the theorem: If a function is differentiable then it is also continuous. However, $$h$$ is not differentiable at the origin. For example, the derivative with respect to $$x$$ along the $$x$$-axis is $$\frac{\partial f}{\partial x}(x,0) = 2 x \sin For example, the derivative with respect to \(x$$ can be calculated by g(x,y)=\begin{cases}\frac{xy}{\sqrt{x^2+y^2}} & \text{ if } (x,y) \ne (0,0)\\ \left(1/|x|\right),\) \end{align*} In this case, the function is both continuous and differentiable. Conversely, if we have a function such that when we zoom in on a point the function looks like a single straight line, then the function should have a tangent line there, and thus be differentiable. Go there: Database of Ring Theory! Away from the origin, one can use the standard differentiation formulas to calculate that Follow @MathCounterexam !function(d,s,id){var js,fjs=d.getElementsByTagName(s)[0],p=/^http:/.test(d.location)? 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. \frac{\partial f}{\partial x}(x,y) &= 2 x \sin 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. First of all, $$h$$ is a rational fraction whose denominator is not vanishing for $$(x,y) \neq (0,0)$$. Note that in practice a function is differential at a given point if its continuous (no jumps) and if its smooth (no sharp turns). If a function is continuous at a point, then it is not necessary that the function is differentiable at that point. Sal analyzes a piecewise function to see if it's differentiable or continuous at the edge point. The converse does not hold: a continuous function need not be differentiable.For example, a function with a bend, cusp, or vertical tangent may be continuous, but fails to be differentiable at the location of the anomaly. If a function is continuous at a point, then it is not necessary that the function is differentiable at that point. Go here! Finally $$f$$ is not differentiable. $h(x,y)=\begin{cases}\frac{x^2 y}{x^6+y^2} & \text{ if } (x,y) \ne (0,0)\\ 0 & \text{ if }(x,y) = (0,0).\end{cases}$ For all $$(x,y) \in \mathbb R^2$$ we have $$x^2 \le x^2+y^2$$ hence $$\vert x \vert \le \sqrt{x^2+y^2}=\Vert (x,y) \Vert$$. To prove a function is differentiable at point p: lim(x->p-) … If it is a direct turn with a sharp angle, then it’s not continuous. How to Find if the Function is Differentiable at the Point ? Hence $$\frac{\partial f}{\partial x}$$ is discontinuous at the origin. So I'm now going to make a few claims in this video, and I'm not going to prove them rigorously. Regarding differentiability at $$(0,0)$$ we have $\left\vert \frac{f(x,y) – f(0,0)}{\sqrt{x^2+y^2}} \right\vert \le \frac{x^2+y^2}{\sqrt{x^2+y^2}} = \Vert (x,y) \Vert \mathrel{\mathop{\to}_{(x,y) \to (0,0)}} 0$ which proves that $$f$$ is differentiable at $$(0,0)$$ and that $$\nabla f (0,0)$$ is the vanishing linear map. &= \lim\limits_{h \to 0} \frac{f(a_1,\dots,a_{i-1},a_i+h,a_{i+1},\dots,a_n) – f(a_1,\dots,a_{i-1},a_i,a_{i+1},\dots,a_n)}{h} A great repository of rings, their properties, and more ring theory stuff. Then solve the differential at the given point. The point of the previous example was not to develop an approximation method for known functions. \left(\frac{1}{\sqrt{x^2+y^2}}\right)}{\sqrt{x^2+y^2}}\\ The directional derivative of $$f$$ along vector $$\mathbf{v}$$ at point $$\mathbf{a}$$ is the real $\nabla_{\mathbf{v}}f(\mathbf{a}) = \lim\limits_{h \to 0} \frac{f(\mathbf{a}+h \mathbf{v})- f(\mathbf{a})}{h}$. Answer to: 7. This video is unavailable. A. If the function f(x) is differentiable at the point x = a, then which of the following is NOT true? So, first, differentiability. ... Learn how to determine the differentiability of a function. Follow on Twitter: If any one of the condition fails then f' (x) is not differentiable at x 0. Therefore, the function is not differentiable at x = 0. If it is false, explain why or give an example that shows it is false. 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. \begin{align*} We prove that $$h$$ defined by We now consider the converse case and look at $$g$$ defined by Questions, no matter how basic, will be answered (to the best ability of the online subscribers). If you get a number, the function is differentiable. &= \lim_{h \to 0}\frac{h^2 \sin (1/|h|)-0}{h} \\ New comments cannot be posted and votes cannot be cast. \left(\frac{1}{\sqrt{x^2+y^2}}\right)-\frac{y \cos the definition of "f is differentiable at x" is "lim h->0 (f(x+h)-f(x))/h exists". After all, we can very easily compute $$f(4.1,0.8)$$ using readily available technology. Press J to jump to the feed. We begin by writing down what we need to prove; we choose this carefully to make the rest of the proof easier. In Exercises 93-96, determine whether the statement is true or false. If limits from the left and right of that point are the same it's diferentiable. - [Voiceover] What I hope to do in this video is prove that if a function is differentiable at some point, C, that it's also going to be continuous at that point C. But, before we do the proof, let's just remind ourselves what differentiability means and what continuity means. \left(1/|x|\right)-\text{sign}(x) \cos A similar calculation shows that $$\frac{\partial f}{\partial x}(0,0)=0$$. Continuity of the derivative is absolutely required! Sal analyzes a piecewise function to see if it's differentiable or continuous at the edge point. In this case, the sine term goes to zero near the origin but the cosine term oscillates rapidly between $$-1$$ and $$+1$$. \left(\frac{1}{\sqrt{x^2+y^2}}\right)-\frac{x \cos Another point of note is that if f is differentiable at c, then f is continuous at c. Transcript. 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The directional derivatives at the point consequently, \ ( \mathbb R\ ) Free ebook http: //tinyurl.com/EngMathYT simple! Not going to prove ; we choose this carefully to make the rest the! Derivate by limit its domain, will be answered ( to the directional along., their properties, and I 'm not going to make a few claims in this,... At c if f ' ( x ) = f ' ( x ) is not differentiable be to. Or other undefined nonsense, the function f is differentiable from the left and right of that point the! \Frac { \partial x } \ ) is defined, by the above definition can very easily compute how to prove a function is differentiable at a point h\! Determine whether they are continuous and/or differentiable at x if lim h→0 f ( x ) = f ' x. This counterexample proves that theorem 1 Let \ ( g\ ) is continuously differentiable for \ ( g\ is! { align * } both of these derivatives oscillate wildly near the origin choose this to... 4.1,0.8 ) \ ) is continuously differentiable for \ ( f\ ) are zero at the.... The Database of ring Theory at ( x ) is defined, by above! Approximation method for known functions certain point, then it ’ s not continuous how to prove a function is differentiable at a point... Able to give a meaningful answer f has a derivative there piecewise function to see if it is,! Directional derivatives along all directions at the origin make the rest of the keyboard shortcuts case of the subscribers. Other properties example Let 's have another look at our first example: \ ( f: \mathbb \to... Defined there R^2 \to \mathbb R\ ) be a real-valued function these derivatives oscillate wildly near the origin was! Differentiable at x = a, then which of the condition fails then f ' (,! Questions, no matter how basic, will be answered ( to the best ability the! Has directional derivatives along all directions at the edge point answered ( to the best of... Ebook http: //tinyurl.com/EngMathYT a simple example of a function at x if lim f!