what is the relationship beteen tangent lines and derivaties?
924809
You can take the deriviative of a function (say f ' (x)) and substitute for x the x-coordinate of a point somewhere on f(x) to get the slope of the tangent line that passes through that point on the curve.\n\nThe tangent line is the best approximation of the curve it is tangent to at the point of tangency.\n\nAlso, you can use point-slope form ( y-yo = m (x-xo) ) to find the equation of the tangent line.
771497
No - the slope is going to be a number, not an equation.\n\nd/dx (2x^2) = 4x\n\nAt x = -2, the slope will be -8.\n\nThe equation of the tangent LINE would be\n\ny + 8 = -8(x + 2)\n\ny = -8x - 24, as you have it.\n\nAnd no - getting the terminology right is not a minor point.
803550
With respect to what?\n\nd/dx xy = y + x dy/dx\n\nd/dy xy = x + y dx/dy\n\nd/dt xy = x dy/dt + y dx/dt\n\nIt's just the Product Rule.
392390
it means you have to derive it twice ,which means you have to derive its second order derivative ,for example derivig the distance formula gives you the speed,then deriving the speed which is the 1st derivative of distance gives you the acceleration.
921398
I just taught this section today to my calc students.  You start by finding the first and second derivative.  I recommend looking in your book for an example.  Your book probably has something like this towards the front of the section.  It would be too long winded to write everything down here.  If you don't have your book on you, the web page below has problems that you can do and then check your answer.\n\nhttp://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/graphingdirectory/Graphing.html
858871
let's say its da/dx means that you are taking derivates of a respect to x.
702411
Rewrite equation as:\ny = x - x^-1\n\nDifferentiate:\ny' = 1 + x^-2\nAt (1,0) the gradient is:\ny' = 1 + 1 = 2\n\nSo equation of tangent is:\ny-0=2(x-1)\ny=2x-2
761290
well first you have to get the first derivative\n\nso: the 1st derivative= t*cost+1*sint\n                                  = cost^2+sint\nthen you take the derivative of that\nso: the 2nd derivative= 2cost*-sint+cost\n                                   = -2sintcost+cost
1101719
Step 1: Find the slope of the line x-2y = 2.  Call it m. \n\nStep 2: Find the derivative f'(x) of the function f(x) = (x-1)/(x+1).\n\nStep 3: Find all values of x where f'(x) = m.\n\nStep 4: For each such value, find its corresponding y-coordinate by setting y=f(x), and then find the equation of the line that goes through that point with slope m.\n\nGood luck!
832569
df/dx=(df/dg)*(dg/dx)=x’=1, dg/dx=(dg/df)*(df/dx)=x’=1; both are const\nIt means g’(b)=1
794869
Here's a good website that discuss implicit differentiation with some example problems.\n\nHope this helps
787521
Example:\nIf z=f(u,v),\nu=g(x,y) and\nv=h(x,y), then\n(all derivatives are supposed to be partial derivatives)\ndz/dx=dz/du*du/dx+dz/dv*dv/dx\ndz/dy=dz/du*du/dy+dz/dv*dv/dy\n\nAlternatively, you can set of the matrices\n[[dz/du dz/dv]]\nand\n[[ du/dx du/dy]\n[dv/dx dv/dy]]\nand get that\n[[dz/dx dz/dy]]\nis the product of the two matrices.
795130
y = tan(πx - 1)\nlet u = πx - 1\nnow, dy/dx = (dy/du)(du/dx)\nd(tan(u)/du = sec^2(u)\ndu/dx = π, so \ndy/dx = πsec^2(u)\ndy/dx = πsec^2(πx - 1)
85418
dx/dt+ax=b :P\n\n\nmathematica had some online find-the-derivitive program, last I remember
763677
the derivative of a function is the slope of the tangent line at any given point or the rate of change at any given point.  the slope of the tangent line of any point on f(x) where x=a is equal to  lim x->a of (f(x)-f(a))/(x-a)
735467
I don't know what it means to "integrate a derivitive [sic]", but, for a continuous function f(x), it does follow that the slope of the line tangent at point (a,b) would be equal to f'(a).
646764
Differentiation is a branch of calculus which determains the gradiant at a specific point on a graph. A straight line graph is easy enough, because between two points on the line, you can divide the distance vertically (y direction) by the distance horizontally (x direction). For example, a graph that has a straight line that is 45 degrees from the x-axis (which, incidently would be 45 degrees from the y-axis as well) would have a gradiant of 1.\n\nIf you wanted to determain the gradiant at a point on a graph which is curved though, you cannot just simply divide the x distance by the y distance, because it is a curve. This is where differentiation comes in. Imagine two points on the curved line, determain the gradiant of that. Then bring the distance of the two points closer together and determain the gradiant of that. As the distance of the two points tends to zero, you would get closer and closer to the gradient of the first point.\n\nAs a generalised equation, you could determain the x and y coordinates of the first point, and the coordinates of the other point by the coordinates of the first point plus the distance between them (e.g. x+dx, x+dy). y would be a function of x in a general equation so you would have the first point coordinates as x, f(x).\n\nNot easy to explain this without diagrams, but I have added a link which you can visually see this.