What is the derivative at x=5?
917622
In order to take h'(x), you need to perform the product rule on the first term because it is multiplied, thus:\n\nh'(x) = d(sin[pi*x])/dx[f(x)] + sin[pi*x][f'(x)] + g'(x)\n\nSimplifying, you get:\n\nh'(x) = pi*cos[pi*x][f(x)] + sin[pi*x][f'(x)] + g'(x)\n\nTherefore, h'(5) = (pi)cos[5pi](f(5)) + sin[5pi](f'(5)) + g'(5)\nh'(5) = pi(-1)(3) + (0)(1) + 0\n\nTherefore, h'(5) = -3*pi\n\n-----------\n\nHope this helps
590233
Hint:\nDo a simple substitution u = x-5\nWhich means x = u + 5\n\nYou should get\n\n(2/5)(x-5)^(5/2) + 2(x-5)^(3/2) + C
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~ ♥ ~
1121843
the given function is not differentiable at x=0\n\ntherefore,   f(x) = x   for x>0        dy/dx = 1\n                           f(x) = -x  for x<0        dy/dx = -1
1105592
y+dy=5(x+dx)^2 +1\n  y+dy=5(x² +2xdx +dx²) +1\n    y+dy=5x²+1 +5(2xdx+dx²)\n       dy=10 xdx as dx² is very small\n          dy/dx= 10x
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
845625
y = -x^2 + 6x - 5.5\ndy/dx = -2x + 6\nInsist dy/dx = 0 to find local maximum.\n0 = -2x + 6\n2x = 6\nx = 3\ny = -(3)^2 + 6(3) - 5.5\ny = -9 + 18 - 5.5\ny = 3.5\n\nThe vertex is at (3, -3.5).\n\nIf you have a polynomial equation of the form,\ny = a x^n + b x^(n-1) + ...\nThen the first derivative is\ndy/dx = an x^(n-1) + b(n-1) x^(n-2) + ...\nWhen dy/dx is zero, the graph of y(x) has either a maximum or a minimum.  In general, dy/dx is the slope of the line tangent to the curve of y(x) at x.\n\nThe neat thing about using differential calculus to find maxima and minima is it works for just about all (differentiable) functions, including polynomials and trigonometric functions.  Once you know this trick, you can nearly always find those local maxes and mins. Duke95's answer is a special case that can be derived from the more general differential calculus method.
972569
By definition of the derivative\ng'(x) = lim h=>0 [g(x + h) - g(x)] /h =  lim h=>0 [3xh + h^2]/h = \nlim h=>0 [3x + h] = 3x
945648
Take the derivative of f :\nf'(x) = 5x^4 (x+7)^4 + 4 (x+7)^3 x^5\nf'(x)= x^4 (x+7)^3 [ 5 (x+7) + 4x]\nf'(x)= x^4 (x+7)^3 [ 9x + 35]\n\nSo the derivative is equal to 0 for the values :\nx= -7\nx= - 35/ 9\nx = 0\n\nthe function is increasing in [-11;-7],\nthe function is increasing in [-35/9;+12],
921060
In general if you have x^n, the derivative is n * x^(n-1).\n\nFor x² the answer is 2x^1 or just 2x.\n\ny = x²\ndy/dx = 2x
199440
find the tangent of the equation x+y=5
945033
Derivative is y=cos x -5. Substituting in 0 gives you the slope -4. Now plug in the coordinates (0,3) to get the line, which is y=-4x+3
716213
Sure,\n\nRemember it is x units, therefore although you are correct that taking the derivative and setting it equal to 0 yields a value for x of just over 5.5 units you have to round it up to 6 units, under the assumption that x is in whole units per day. That's alright for a daily figure but 6 is incorrect if you are producing amounts averaged over longer periods of time. In short: you are correct and the answer in the book is stupid! Technically though, if C'(x) is your derivative function then you need to establish that C'(6) is closer to 0 than C'(5). C'(5) = -0.4 and C'(6) = 0.274, so C'(6) is closer to 0 and is the correct answer to a stupid question. lol.\n\nYou do have the correct cost function C(x)\n\nC'(x) = 1.2 - 200/x^3\n\nC'(x) = 0 when x^3 = 200/1.2 = 166 2/3\n\nx = 5.503
506639
The first derivitive of f(x) = 5/x^3 - 3/x^1/3 is...\n\nf'(x) = -15/x^4 + 1/x^(4/3)\n\nHope this helps. I used the product rule. Remember, you can write the term 5/x^3 and 5*x^-3. I find it simpler to rearrange the equation and use the product rule to get at the first derivitive.
676902
Incomplete question. Are you looking for the anti-derivative of x, a, or b???\n\nput dx, da, or db at the end of the equation so we'll know.
974331
= ln x / x^5\nln x -> inf\nx^5 -> inf\ndifferentiate both sides (Lagrange)\n\n=> 1/x  / 5x^4\n= 1/ (5x^5)\n=> 0 \n\nOr ln x increases much slower as x gets bigger, whereas x^5 increases faster, (exponentially) as it gets bigger. polynomial wins over logarithm. \n\nThe limit is 0.
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