Implicit Differentiation?
956578
x^2+3x-y=c\n2x +3 -y'=0\ny'= 2x +3\n\n(remember y^0=1)!
599545
No. Look at f(x) = |x|. It is everywhere continuous but not differentiable with respect to x at x=0.\n\nThis is a stupid question, and you're stupid for asking it.
1001167
The solution to your question is a bit lengthy and it wont be easy typing that here. However I can tell you the guidelines to arrive at that.\n(1)Write down the equation\n(2) take differentiation wrt x on both sides.RHS will be 1 here.LHS will have to be diff..you must know how to diff e^(y/x).Use the diff rule for products.\n(3) take diff.again of the above obtained equation.you will get alot of terms then.try getting from that the form you want.
547629
assuming u is an unknown variable y=0
544525
You need to show that there are no horizontal tangents.\nTake the derivative, set it equal to zero, and show that there are no real values for which the derivative equals zero.
985531
To solve by implicit differentiation, you need to solve for 'dy/dx'.  \n\nd[x^3]/dx + d[y^3]/dx = d[3xy]/dx.  Taking the derivative, you get:\n\n3x^2 + 3y^2(dy/dx) = 3y + 3x(dy/dx) [used the chain rule on the right side].  Isolating dy/dx, you get:\n\ndy/dx = (3x^2 - 3y)/(3x - 3y^2).  Factoring out a 3 in both the numerator and denominator, you get:\n\ndy/dx = (x^2 - y)/(x - y^2)\n\n-------\n\nNow to solve for the horizontal tangent line, you need to determine when the slope (i.e. dy/dx) = 0.  That is the definition of a horizontal line.\n\nTo solve dy/dx = 0, set the numerator = 0.  Thus, you get:\n\nx^2 = y\n\nAs you can see, there are an infinite number of ordered pairs (x,y) that satisfy this equation.  Therefore, there are an infinite number of horizontal tangent lines that can be made.  But make sure the denominator does not = 0, or the slope is undefined (vertical asymptote), or when x = y^2.  An ordered pair that fits the above equation is (2,4).  Therefore, y = 4 is a horizontal asymptote.  y = 9 is a horizontal asymptote at (3,9).  \n\nThe list goes on and on and on.....\n\n--------\n\nHope this helps
881328
this requires way more than anyone can do here, but, i know of a cool java program that will do it for you: \n\nhttp://www.math.rutgers.edu/courses/ODE/sherod/phase-local.html\n\nyou can even set your initial conditions and print it out!
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.
48881
Anytime you have a problem that is continuous, as opposed to discrete, you will need to use either integration or differentiation to characterize the solution.  That means to say that every problem that involves time in one way or another will necessitate the use of calculus to find an answer at some instant in time.  If the problem is such that you can approximate the answer, then you can discretize the parameters as small slices of time and come up with an approximate answer - without using calculus.
761613
Using implicit differentiation, you get:\n\n-csc^2(y)(dy/dx) = 1 - dy/dx\n-csc^2(y)(dy/dx) + dy/dx = 1\n(dy/dx)[1 - csc^2(y)] = 1\n\nTherefore, dy/dx = 1/[1 - csc^2(y)]\n\nNow plug in your ordered pair to get dy/dx.  Remember csc (y) = 1/sin(y):\n\ndy/dx = 1/[1-1], which is undefined.  Therefore, you cannot write a tangent line for this ordered pair (there is a vertical asymptote at this point)\n\n------------\n\nHope this helps
425296
Instantaneous rate of change of a function (7 words),\n\nOR,\n\nSlope of the tangent line (5 words). :P
218622
differentiate wrt x.. dy/dx = 2x + 3 = 0\nx = -3/2 is the vertex (-3/2. -49/4)\n\nif you don't know differentiation,\nexpress in terms of (x-a)^2 + b = y, \ny = (x - -3/2)^2 + 31/4\n(-3/2, -49/4) is the vertex. (Value of y is found by substituting value of x in equation)\n\nx intercept is found by placing y=0, (-5,0) and (2,0)\ny intercept is found by placing x=0, (0,-10)
1122081
It's called the integration by parts formula.\nInt u dv=uv- int v du\n\nBoth u and v need to be absolutely continuous for this to work. A continous derivative for both is plenty.
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.
43939
is the above question in first degreee derivative??
1116918
In Calculus of Varations deals with minimization problems the most elementary of which is the Brachistochrone.  This is the problem of the shape of a wire which provides the fastest path between two points for a marble sliding down it.  I don't know what applications of differentiation is.
303213
Expressions such as yours is called a power tower. They are differentiable by writing them as exponentials but they are not integrable in closed form. You can find the series form at the source link however.