Can a differentiable function (everywhere) fail to have a tangent?
603169
phyzikzteacher is right, as well as the next answer after his.  Cusps and corners are notorious points where a function fails to have a SINGLE derivative.  In fact, cusps and corners have an infinite number of tangent values at those points.  The function, however, is differentiable, just that you get a zero in the denominator usually at the x value of the cusp/corner.
793950
order 1 because the iterate does not involve any power of the derivative at xn to be known in order to obtain the next iterate
794869
Here's a good website that discuss implicit differentiation with some example problems.\n\nHope this helps
765864
I'm unclear with the way you stated the problem. Am I looking for a implicit derivative. I'll check back to solve it.
793364
To show that it doesn't exist, look at the path where x=0 and y moves to zero.  The problem reduces to:\n\nsin(y)/y^2\n\nYou can use L'Hospital's rule to show that this limit does not exist.
355882
The relationship is defined in one dimension as f'(x) being the slope of the tangent line to the graph of f(x) at the point x. In symbols, f'(x) is the limit as h approaches 0 of the slope of a secant to the graph passing through x: [f(x + h) - f(x)]/h.\nf'(x) being the slope of the tangent line to the graph of f at x has its advantages. When f'(x) is 0, we know the tangent line is horizontal there, meaning the graph is either at a minimum, a maximum, or a temporary flat area. Higher derivatives (treating the derivative as a function itself) can tell us which of the above it is.\nDerivatives are most concrete in physical examples. If x(t) is the position of an object at time t, then the velocity of the object at time t can be idealized as the derivative of x(t) at t (instead of the average velocity, which is the slope of a secant line to the graph of x(t)). Similarly, the acceleration at time t is given by the second derivative.\nIn higher dimensions, where x = (x1, x2, ..., xn) is a point and f(x) = (f1(x), f2(x), ..., fn(x)) is also a point, this is generalized to f'(x) being the unique linear function that vanishes with the change in f at x. In symbols, the limit as ||h|| vanishes of [f(x + h) - f(x) - f'(h)]/||h|| vanishes. This coincides with the definition in 1 dimension above. In higher dimensions, f'(x) corresponds to the basis of a tangent space at x, as it did in 1 dimension.\nIf f is a map from R^2 into R, f'(x) is a linear map from R^2 into R, which is a plane. This plane is a tangent space to the graph of f at x. Higher-dimensional tangent spaces are not as visualizable.
824813
f(x) = x^m(1-x)^n\nm zeros at  x = 0 and n zeros at x = 1 \nSo x-axis is tangential to the curve at x = 0 and x = 1\n\nf'(x) = mx^(m - 1)(1 - x)^n + x^m * -n(1 - x)^(n - 1)\n= x^(m - 1) * (1-x)^(n - 1) * (m(1 - x) - nx)\n=  x^(m - 1) * (1-x)^(n - 1) * [m - (m + n)x]\n= 0 for stationary points.\n\nSo (m - 1) zeros at x = 0, (n - 1) zeros at x = 1 and a zero at x = m / (m + n)\n\nNow there has to be a turning point between x = 0 and x = 1 as f(x) is a polynomial and therefore continuous and differentiable for all x\n\nSo examine the possibilities\n... Domain → ... x < 0 ... 0<x<1 ... x>1 \nm .... odd    \nn ..... odd    \nx^m ..................... - .......... + ........... + ... Stationary inflexion at (0, 0)\n(1 - x)^n .............. + .......... + ........... - ... Local maximum at (m/(m+n), f(m/(m+n))\nf(x) ....................... - .......... + ........... - ... Stationary inflexion at (1, 0)\n\n\n... Domain → ... x < 0 ... 0<x<1 ... x>1 \nm .... odd    \nn ..... odd    \nx^m ..................... - .......... + ........... + ... Stationary inflexion at (0, 0)\n(1 - x)^n .............. + .......... + ........... + ... Local maximum at (m/(m+n), f(m/(m+n))\nf(x) ...................... - .......... + ........... + ... Local minimum at (1, 0)\n\n\n... Domain → ... x < 0 ... 0<x<1 ... x>1 \nm .... odd    \nn ..... odd    \nx^m ..................... + .......... + ........... + ... Local minimum at (0, 0)\n(1 - x)^n ............... + .......... + ........... - ... Local maximum at (m/(m+n), f(m/(m+n))\nf(x) ....................... + .......... + ........... - ... Stationary inflexion at (1, 0)\n\n\n... Domain → ... x < 0 ... 0<x<1 ... x>1 \nm .... odd    \nn ..... odd    \nx^m .................... + .......... + ........... + ... Local minimum at (0, 0)\n(1 - x)^n .............. + .......... + ........... + ... Local maximum at (m/(m+n), f(m/(m+n))\nf(x) ...................... + .......... + ........... + ... Local minimum at (1, 0)
566883
First, since 1/x is not defined for x = 0, before you can show  anything about your functions, you have to define values for g(0) and f(0). I'm going to assume that either the original statement of the problem had that or that you are supposed to find the value that makes the problem solvable.\n\nTo be true you have to define the problems with f(x) = 0 and g(x) = 0\n\n\nFor the continutiy of f(x) given epsilon > 0 find delta > 0 such that |f(x) - f(0)| < epsilon for all |x-0| < delta.\n\n|f(x) - f(0)| = |xsin(1/x) - 0| < |x| = |x - 0|\n\nGiven epsilon, pick delta = epsilon.\nIf |x-0| < delta, then |f(x) - f(0)| < epsilon.\n\n\nAs for the differentiability of g(x),\n\nf(0+h) - f(0)) = (0+h)^2sin(1/(0+h)) - 0\n\nlim(x->0)(f(0+h) - f(0))/h = \nlim(x->0)((0+h)^2sin(1/(0+h)) - 0)/h = \nlim(x->0) h^2(sin(1/h)/h = lim hsin(1/h) = 0\n\nSo the limit exists and is 0.
507403
Behind all the nomenclature, this is just a simple minimization problem. You want to find local minima for A(q). Thus, take the derivative of A(q) to look for critical points (places where the graph is locally flat). These will be places where the tangent line has slope 0; so set the derivative equal to 0 and solve for q. You will get two possible choices. Use the second derivative to see what the behavior of the first derivative is at each point (increasing or decreasing). An increasing first derivative around a critical point (where the derivative is zero) implies that the slope locally changed from negative values to positive values. Visualize this to see that it is a minimum point for the graph of A(q). The q that gives you the positive second derivative and has 0 first derivative is thus q of the Wilson lot size formula.\nFor b, simply replace k with k + bq in A(q) and do a similar analysis as above.\n\nPS. The answerer above has forgotten the negative value of q that also satisfies the first equation. Although a negative q does not correspond to the problem, it is unrigorous to say that there is only one answer to the equation. It should at least be explained as being unphysical.
271844
If you need the tangent, take the derivative of the function at the specific point, which will give you the slope. Since you have taken the derivative at some specific point (a,f(a)) you also have a point. Now use the point-slope form to figure out the equation. Then simply get a graph paper and graph it.\n\nExample: f(x)=3tan(2x)\n\nf'(x)=6[sec(2x)]^2 (I assume you can take derivatives.)\n\nYou have a point (a,f(a))\n\nNow use the point-slope form: y-f(a)=f'(x)(x-a)\ny-3tan(2a)=6[sec(2a)]^2*(x-a) is the answer in this case. \n\nOr in function form: y=6[sec(2a)]^2*(x-a)+3tan(2a)\n\nTo graph it, find 2 points on the equation, then draw a straight line between them. That is your tangent.
450251
Yes, it is quite possible. First imagine such a curve and take a tangent line at an interior point. Now move that line up slightly so it intersects the original function at two nearby points. Finally let the line curve very slightly between those intersection points and join the resulting curve to (0,0) and (1,1) smoothly. Specific examples should be easy with this intuition.\n\nI'm guessing that any number of intersection points are possible. Just modify this proceedure between intersection points.
547668
--- An informal definition of a continuous function is a graph that can be drawn without picking up your pencil.\n--- This graph, however, is NOT differentiable at the point x = 0 because it is a cusp, i.e., the slope of the graph from the left is not equal to the slope of the graph from the right.\n--- If you take the derivative, you get -(2/3)x^(-1/3). If you substitute in x = 0, you get an undefined derivative. i.e., not differentiable at x = 0.
226178
Just doing the two partial derivatives and setting them to zero is not enough because you cam have a change of inflection (if I'm not mistaken). With one variable only, that would be equivalent to taking the derivative and equaling it to zero for f(x)= x^3. f'(x) = 3x^2 and f'(0)=0. However, x=0 is not a local extreme, it is only a candidate. This is where analisys takes place. One way to verify if this candidate is infact an extreme is to calculate the second derivative. If it is >0, it is a local minimum, if it is <0 it is a local maximum. In this case f''(0) = 0, so it is simply a point of inflection (check this terminology cause I've graduated a long time ago and am quoting from memory).\nFor a two variable function, you have to repeat this for y. So a candidate will be a point (xi, yi) in which both partial derivatives are zero and both second derivatives have the *same* sign (otherwise you have a point of saddle: I dont know if this is the correct terminology in English) and both different from zero.
442091
The integral of a function can be defined as the limit of those rectangles as they get smaller and smaller, but I think what you are asking is--why is this connected to the anti-derivative?  Why is it that you can take the anti-derivative of a function, plug in at the endpoints, and you get the area under the curve?\n\nThis is a hard question to explain; it's generally called the "Fundamental Theorem of Calculus", and most textbooks prove it in detail.  You can also check out the wikipedia article below, although it's certainly not easy to understand.\n\nIf I could sum it up in a nutshell I would say that the reason why it works is that taking area under a curve is like adding up lots of little pieces, but taking the derivative (slope) is like subtraction...little tiny pieces at a time.  Just as addition and subtraction are opposites, differentiation (taking slopes) and integration (taking areas) are opposites.\n\nIt also helps to play around with the two functions f(x)=1, and f(x)=x.  Sometimes experimenting with these simple functions can help give you a feel for why this crazy connection really works.
851986
Tangents do not exist. Neither does calculus. They are a mass conspiracy by the government. Both Newton and Leibnitz were part of a secret society called the Tangent Obliterators. What really happened when they invented calculus was that Newton and Leibnitz had one too many drinks that evening. And they figured, "Hey why not **** with everybody and invent a new kind of math!"\n\nBut anyway, take the derivative of each function and set them equal to each other, then solve for x.
948694
The anti-derivative of a function f(x) is another function F(x) so that F´(x) = f(x). \n\nIf f(x) = x^n (n<-1 or n>-1) then the anti-derivative is F(x) = (x^(n+1))/(n+1)  + C ... we need to add an arbitrary constant.\n\nIn your case we have:  f(x) = x^1  and  F(x) = (x^2)/2 + C
364736
This statement can be proven by proving the contrapositive: If f:R -> R is neither strictly increasing nor strictly decreasing, then f is not a 1-1 differentiable function.\nIf f is differentiable, then f being neither strictly decreasing nor strictly increasing is equivalent to f being a constant function over some nondegenerate interval in its domain. Thus f cannot be 1-1.\nIf f is 1-1, then the condition implies that f is not continuous, and thus is not differentiable.