Two Questions: 1) Write the equation for the tangent line to the graph of f(x)=e^(x) ln(x^(2) + e) at X=0?
911982
slope = f'(0),\nf'(x)= e^x 1/(x^2 +3) 2x + e^x ln( x^2 +e)\nf'(0)= e^0 1/3 (2(0)) + e^0 ln(e)\n= 1\n\nnow y=f(0)=e^0 ln( e) =1\nso the eq. of the tangent line is:\n\ny-1= x\ny=x+1\n\n\nfor question 2:\ne^{sin(0.001)} is very similar to e^0.001 which in turn is approx 1\ns
897760
The slope of the tangent is 2x, so the slope of the normal line is -1/(2x), not -1/2 x, which you used later.\nIf you continue of, you get -1/(2a), so the equation is y = -1/(2a) * x + b.\nSo a^2 = -1/(2a) * a + b = -1/2 + b.\nThus b = a^2 + 1/2.\n\nIn this case, theres no dividing by a, so you can just substitute in a=0 if you wanted (that was a good hint that your answer was wrong).
517319
Using the definition of the derivative.\nf'(x) = lim(as h->0) of [f(x+h)-f(x)]/h\n\nLet f(x) = mx + b\nf(x+h) = m(x+h) + b\n\nf(x+h) - f(x)\n= m(x+h) + b - (mx+b)\n= mx + mh + b - mx - b\n= mh\n\n[f(x+h)-f(x)]/h \n= mh/h\n= m\nTherefore,\nf'(x) = lim(as h->0) of [f(x+h)-f(x)]/h\n= lim(as h->0) of m\n= m
821550
if f(x) = ∫[t = x to 0] 1/ (1 + t + t²) dt\n\nf'(x) = -1/ (1 + x + x²)\n\nf''(x) = (1 + 2x)/(1 + x + x²)²\n\nFor concave UP f''(x) > 0\n\nSo 1 + 2x > 0 (as (1 + x + x²)² > 0 for all x)\n\nie x < -1/2
493069
First :\nSince f is continuous over its domain (R),\ndf/dx = 5 x^4 +2 > 0 for every x\nSo f is an increasing function over its domain\nHence for every value of y their exist a unique value of x\nHence f has an inverse and one-one function\n3- Use function graph to graph f-1 as you mentioned\n(df^-1/dx) (2)= the slope of the tangent to f-1 at x=2\n6- (1,2)\n7- 7\n8- 1/7
801587
Are you in pre-calc or calculus yet?  If so, you could find the vertex by taking the derivative of the function and finding where it equals zero (the derivative = the slope of the tangent to the curve, and the tangent to the curve = 0 at the vertex).\n\nf(x) = 5x² - 25x + 5\n\nf '(x) = 10x - 25\n\nf '(x) = 0 when x = 2.5\n\nSo, now find f(2.5) to determine the y-coordinate of your tangent:\n\nf(2.5) = 5(2.5)² - 25(2.5) + 5\n= 31.25 - 62.5 + 5\n= -26.25\n\nSo your vertex is at the point (2.5,-26.25)\n\n~ ♥ ~
632327
Write f(x,y) as x^2*[y^2/(x^2+y^2)]. The second factor is always <=1, so f(x,y) goes to 0 as (x,y) goes to (0,0). The first answerer is WRONG. As for the second answerer, just knowing the limit is 0 along every line is not enough to conclude that the limit exists. It is possible for limits along lines to be 0, but to be otherwise along parabolas.\n\nIf f(0,0) is defined to be 0, then this function has partial derivatives at (0,0) that are both 0. In fact, f is differentiable at (0,0) in that case. The derivative is the 0 functional.
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.
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.
51150
UGH!!!  Thx for the effort (I was the one to post that question, as a part of a solution to another problem I found here on Yahoo! Answers.)  Incidentally, I made a mistake when simplifying the other problem and it should be xln(2)-2ln(x)-ln(86400) = 0\n\nI don't know any way to get rid of the ln(x) unless maybe it's using calculus techniques?  I'm just glad this problem is making other ppl trouble about it as much as I did :-).
715464
y' = (sinx-cosx) / (sinx+cosx)^2\n\nTo find the tangent at (0,1) you need its coefficient, which is y'(0) = -1.\n\nYour tangent should be y - 1 = -1.(x - 0)\ny = -x + 1
752959
Problem 1: Find an exponential function of the form f(x)=ba^x+c with y-intercept 2, horizontal asymptote y=-2, that passes through the point P(1,4). \n\nYou find the y-intercept by replacing x with zero, which gives f(x) [which is y on the graph] = 2, so b + c = 2 (since a^0 = 1)\n\nThe horizontal asymptote is y = c, so c = -2.  That means b must be [you figure it out]\nNow it is y = 4a^x - 2.  This must be true when x=1 and y=4.\nSo 4 = [you figure it out; answer deleted due to complaint from below]\n\n\nProb. 3.  What is the amount in the account if $500 is invested for 10 years at the annual rate of 5% compounded continuously? \n  Just plug in the values.\nA = 500 e^(.05 x 10)\n\n[answer deleted]
347784
Taking natural logarithm to both sides\nLn(y)= Ln(x) * Ln(x) = (Ln(x))^2\nDifferentiating\n(1/y )dy/dx = (2/x) * (Ln(x))\ndy/dx = (2y/x) * (Ln(x))\nat the point (e,e),\ndy/dx = (2e /e) * Ln(e)\nso the slope = 2
1085682
Ln(e^x+1) = 0 by log laws this equals X+1 Ln e = 0 and Ln e cancels therefore X+1 = 0, x = -1\n\nIf the equation does not = 0 then you will be just left with ( x + 1)
226683
The equation of the tangent at any point is f'(x)=3*(x^2).\n\nAt x=2, f'(2)=12, f(2)=8.\n\nSo the line has slope 12 and passes through the point (2,8).\n\ny - 8 = 12 * (x - 2) is the equation of the tangent line.
1130920
You can't solve this using traditional techniques since you'll either have a logarithm or an exponential on one of the x variables.  The people claiming Taylor series and techniques like Newton-Raphson approximation are most likely on the right track.  I would use a CAD program, or graph the function y(x)=(1+x)^2-2^x and find the zeroes that way by zooming and such (on at TI-83 or similar).\n\nIn this case, I'll use MathCAD.  The solutions are:\nx={-1.57862, 0, 5.31972} (two of these are approximate, by the way)
781707
Find the first derivative and set it equal to zero.\nf'(x) = 6x^2 - 30x + 24\n6x^2 - 30x + 24 = 0\nDivide everything by 6\nx^2 - 5x + 4 = 0\nFactor\n(x-1)(x-4) = 0\nx = 1 or x = 4\nTo test if these are max or min's, we can use the second derivative.\nf''(x) = 12x - 30\nFor x = 1, f''(1) = 12(1) - 30 = -18 < 0\nThis means f(x) is concave down at x = 1; hence, x = 1 is a local max.\nFor x = 4, f''(4) = 12(4) - 30 = 18 > 0\nThis means f(x) is concave up at x = 4; thus, x = 4 is a local min.\n\nIf you are looking for the points, then you will need to substitute x = 1 and x = 4 into f(x) to find the corresponding y value for each.