how do get the derivative of  f(x)= x^2+2x?
139329
calculus, fun stuff. To get the derivative, you need to drop the exponent then subtract one from the explonent. SO the X^2 would be 2x and 2x would just be 2 .    Final answer is 2x + 2.   Hm.. explained that poorly. In general, to find the derivative, Take the exponent on the X, drop it in front so its multipluing, than subract one from the exponent and have that number in the old exponents place (3x^3 would turn into 9x^2, 4x^2 would be just 8x) any number with X^1 is just the number, and just a number with no X is 0.
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
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!
183101
You can use the product rule the way it is written right now, or you can use foil and then take the derivative the regular way.  I think the second would be less work.\n(x+1)(x+3)\n= x^2 + 4x + 3\nThe derivative is 2x + 4\n\nIf you use the product rule, you would have\n(x+1)(x+3)' + (x+1)'(x+3)\n= (x+1)(1) + (1)(x+3)\n= x+1 + x+3\n= 2x + 4
919994
Find the 2nd derivative and find the max.
689918
use  power rule and product rule\nf'(x)=[1/5]*5*x^4 + 2x(x^2-x-9)+ \n(x^2+9)(2x-1) + 0=\n\nx^4 + 2x^3 - 2x^2 -18x + 2x^3 - x^2 +18x -9 =\n\nx^4 + 4x^3 - 3x^2 -9
1111738
the best way to appr this problem is to apply the first principle i.e.\nf'(x)= Lt     [f(x+h)-f(x)]/h\n       h->0\nnow putting f(x)=e^x we have\nf'(x)= Lt    [e^(x+h)-e^x]/h\n       h->0\n     =Lt    [e^x(e^h-1)]/h\n       h->0\n\nnow (e^h-1)/h where h tends to 0 is 1\nhence f'(x) = e^x\ntherefore the derivative of e^x is e^x
888297
You can of course use a brute force approach to get the derivative but a little finesse often can simplify things.  You have:\n\ny=(sin(x)/1-cos(x))^2 = (sin(x)^2)/(1-cos(x))^2\n\nSince sin(x)^2 = 1 - cos(x)^2 it follows that:\n\ny = (1 - cos(x)^2)/(1 - cos(x))^2\n\nBy doing a little algebra up fornt, the final result is quite a bit simpler than some of the others given.\nFactoring the numerator:\n\ny = (1 - cos(x))*(1 + cos(x))/(1- cos(x))^2 = (1 + cos(x))/(1 - cos(x)) \n\nNow your function is of the form\n\ny = (1 + u)/(1 - u) where u = cos(x)\n\ndy/dx = dy/du * du/dx = (1/(1 - u) + (1 + u)/(1 - u)^2)(-sin(x))\n\ndy/dx = -2*sin(x)/(1 - u)^2 = -2*sin(x)/(1 - cos(x))^2\n\nBy doing a little algebra up front, the final result is quite a bit sinpler than those above.
953139
f'(x)=(4/3)x^(1/3) +(4/3)x^(-2/3)
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
745843
Use the quotient rule\nThe derivative is\n[(Q)(25000 +150 + 3Q)' - (25000 + 150 + 3Q)(Q)'] / (Q)^2\n= [(Q)(3) - (25000 + 150 + 3Q)(1)] / Q^2\n= [3Q - 25000 - 150 - 3Q] / Q^2\n= [-25000 - 150]/ Q^2\n= -25150/Q^2\nIf you want (and I would recommend this), then you can combine 25000 and 150 to get 25150.
180190
let f(x) = (x^2) sin(x+2)\nthe derivitive of f(x) [ f'(x) = 2x sin(x+2) + (x^2) cos(x+2)]\n\nThis is done by taking the product rule {g(x)h(x)}' = g'(x)h(x) + g(x)h'(x)( the derivitive of the first times the second plus the derivitive of the second times the first) \nwhere g(x) = x^2 and h(x) = sin(x+2)
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.
892157
wait 1 sec.\nOk this is reasonably difficult, you have to use the chain rule and the quotient rule.\nUsing the chain rule:\nderivative = [ (1/2) * ( (2x + 3) / (7x+4) ) ^ (-1/2) ]  *  [d/dx (2x+3) / (7x+4) ]\n\nSo now you to use the quotient rule to find:\nd/dx (2x+3) / (7x+4) = (14x+8 - 14x + 21) / (7x+4)^2\n= 29 / (7x+4)^2\n\nSo the derivative is [1/2 * ((2x+3)/(7x+4))^(-1/2)]*[29 / (7x+4)^2]\n\n\nBy the way i got the thumbs down before I wrote all this up, so it doesn't mean it's wrong.
1128734
I'm going to make a wild guess that you are trying to check an answer you got with the answer in the back of the book. I'm also going to assume that you were supposed to take a derivative of\n(x^3/(a-x))^(1/2).\nIn that case, you got the wrong answer because you didn't do the chain rule. You still need to multiply what you have by the derivative of the thing inside, i.e. the derivative of x^3/(a-x).\n\nIf I'm making a bad assumption and this didn't come from a differentiation problem, please ignore my answer.
758976
The total differential is [∂F/∂x]dx +[ ∂F/∂y]dy\n\n∂F/∂x = 4x^3*y^-1\n∂F/∂y = x^4*(-y^-2) = -x^4y^-2
750604
As Stuart has said above, you need to use the product rule to work this out.  The product rule states:\n\nthe derivative of f(x)g(x) is f(x)g'(x) + f'(x)g(x)\n\nSo, when you differentiate sinx wrtx you get cos and when you differentiate 2x wrtx you get 2.  Plugging these into the formula above you get:\n\n(f(x)g(x))'=2x(cosx)+2(sinx)\n                =2xcosx + 2sinx\n\nHope that this is helpful!