Find the second derivative of y = squareroot(2x +3). \n\nI want an answer AND an explanation. If you give me just the answer, then I will get my homework question right, but I will not know how to do it on the final exam. Thank you!
125288
Use the chain rule and the power rule.\n\nIf y = x^a, then y' = a * x^(a-1).\n\nSo, y = (2x+3)^(.5)\ny' = .5 * 2 * (2x+3)^(-.5) = (2x+3)^(-.5)\ny'' = -.5 * 2 * (2x+3)^-1.5 = -(2x+3)^(-1.5)
342922
first get all the y's on one side and the x and y's seperated by rewriting the equation like this: x^3 = (3x-y^2)y. Then you can plug in your x so that you get 2 = (3*2^(1/3)-y^2)y it gets sticky from here. The equation can be writen 2 = ((sqrt(3*2^(1/3)-y)(sqrt(3*2^(1/3 ) + y) ) y which gives you y = 1.5874, .5810, -2.1684. So there you go, I got those values using the solve function on my 89. However the derivative of x^3+y^3=3xy is\n\n(x^2-y)/(x-y^2)=dy/dx\n\nso dy/dx equals 0 whenever y = x^2 and that my friend is your answer.
79831
arctan'(x) = 1/(1+x^2), so differentiating using the good old rules, we get the result (using 'sq' for the square root to avoid yahoo cutting the equation short):\n\nsq(a+x^2)-1\n-------------------------------\n(x*sq(a+x^2)(1+(x-sq(a+x^2))^2\n\nMan, that's ugly. :-)
402841
Eliminate the y to get\n2-x^2=2x+4\nx^2+2x+2=0\nusing quadratic formula\n(-2+-sqrt(4-4*2))/2\nwe see there are no real roots: the two functions do not intersect\nx= -1+-i
571284
Please explain what is ~f.  I have curiosity, because neither inverse multiplicative, complex conjugate or inverse function could do that result...\n\nIt should be interesting!! Please post what is ~f !
516753
uh, find the derivative mathematically and then graph it.\n\nfor example to graph derivative of x^2.\n\nfind the derivative equation f(x)=2x\nand then graph y=2x.\n\nooops, didnt realize it was a picture.\n\nfind the zeroes (a high or low point i.e. tangent is a horizontal line)\nyou should be able to track poitive and negative slopes.  not too hard, you'll figure it out.
504794
For #1, there's no way you can solve that directly.  Note that it can be split up into three fractions, though.\n\n(x^2 + x + 1)/x = x^2/x + x/x + 1/x\n\nNow, we can actually reduce that, to\n\nx + 1 + 1/x\n\nWe solve for the antiderivative using the reverse power rule for derivatives.  That is, antiderivative for x^n is [x^(n+1)]/(n+1).  The exception is 1/x, where the antiderivative is ln|x|.\n\nSo the antiderivative of x+1+1/x is\n\n(x^2) / 2 + x + ln|x|\n\nBut we want the general antiderivative, so we add a constant C.\n\n(x^2)/2 + x + ln|x| + C
519775
1)  \nTo find increasing and decreasing of a function, you need to take the derivative:\n\nsqrt(4-x) can be re-written as (4-x)^(1/2)\n\nTaking the power rule and chain rule, you get (1/2)(4-x)^(-1/2)(-1)\n\nSimplifying, you get -1/(sqrt(4-x)).\n\nTo determine increasing or decreasing of a function, look at the derivative.  If the derivative is positive, the function is increasing and if the derivative if negative, the function is decreasing.  Looking from the derivative, you can conclude that the derivative is only defined for x < 4 because otherwise, you would have a negative square root which has no real value.  Thus, for all x < 4, you can conclude that the derivative is negative (try plugging in a value for x).  \n\nTherefore, the function is decreasing from (-infinity, 4)\n-------------------------------\n2)\nTake the derivative of the function:\n\nf(x) = x^2 - 4x\nf'(x) = 2x - 4\n\nSetting the derivative equal to zero locates your extreme values:\n\n2x - 4 = 0 ---------------> x = 2\n\nNow perform the number line test to see how the function behaves to the right and left of x = 2.  If you plug in a value to the left of 2, let's say x = 1, you get a negative derivative, thus the function is decreasing.  If you plug in a values to the right of 2, let's say x = 3, the derivative is positive.  Therefore, you can conclude from these findings that x = 2 is a minimum.  Solving for the actual point, plug in x = 2 into original function.  You get y = -4.\n\nSince the function extends to infinity in the y-direction for negative and positive infinty of x, (2, - 4) is an ABSOLUTE MINIMUM\n--------------------\n3)\nAgain, take the derivative:\n\nf(x) = x^3 + 5x^2 -13x + 1\nf'(x) = 3x^2 + 10x - 13\n\nNow set the derivative equal to zero to find extreme values:\n\n3x^2 + 10x - 13 = 0\n(x - 1)(3x + 13) = 0\n\nTherefore, you have extreme points at x = 1, -13/3\n\nNow do number line test like above to see if you get max or min.\n\nRemember, \nnegative->positive change = minimum\npositive->negative change = maximum\n\nx = 1\nto the left (like x = 0) ---------> negative\nto the right (like x = 2) -------> positive\n\nThus, at x = 1, you have a minimum.  The exact point is (1,-6)\n\nx = -13/3\nto the left (like x = -5) ---------> positive\nto the right (like x = -4) -------> negative\n\nThus, at x = -13/3, you have a maximum.  Solve for y to get the exact point.\n\nNow determine values at x = -+ infinity \n\nAt x = +infinity, y = +infinity\nAt x = -infinity, y = -infinity.\n\nTherefore, since at x = 1 and x= -13/3 give you y values that are not the extreme values, these are local extremes.  Thus, x = 1 is a local minimum and x = -13/3 is a local maximum\n-----------\n4)\n\nIf you plug in x = 0, you get a positive value thus it is increasing from -infinity to 1.  if you plug in x = 6, you get a positive value, thus the function is increasing from 5 to infinity.  If you plug in a value between 1 and 5 (like 3), you get a negative value, thus the function is decreasing from x = (1,5)\n--------------\n\nHope this helps
531012
Hi again \n\nY = √2x – 3 \n2x – 3 = y ²\nX =( y ² + 3 )/ 2\nf –¹ (x)  =( y ² + 3 )/ 2  =>   f –¹ (x)  = (x ² + 3 )/ 2   &  x ≥ 0 \n\nGood Luck \n try to solve them yourslef again if you got wrong , am sure you can solve them correctly , if you just try your best
100562
The derivative of sinx is just cosx, but the derivative of cosx is -sinx.  Then the derivative of -sinx is -cosx, and the derivative of that is sinx.  So it actually takes four applications of the derivative to get back where you started rather than two.  \n\nIf you're comfortable with the definition of the a derivative you can work it out and see that that's true.  Otherwise, just draw the graphs and look at where the slopes are negative and where they're positive.\n\nHope that helps!\n\nPS.  When you're looking at the page your question is on, you can click "additional information," and add more there rather than making a new question.  Saves you five points. :-)
511398
#1 \n\nIf f(x)= √(x³ + 5x +121) * (x² +x +11) \n\nthen, f(0)= √(121) * (11) \n= 11 * 11\n= 121\n\nf'(0) = -1/2 (See #3 below at ***)\n\n#2 \nIf y is a differentiable function of x, then the slope of the curve xy^2 -2y + 4y^3 = 6 at the point where y = 1 is what?\n\nWhen y = 1\n\nx(1)^2 - 2(1) + 4(1)^3 = 6 \nie x + 2 = 6\nSo x = 4\nSo y^2 + 2xydy/dx -2dy/dx + 12y^2dy/dx = 0\nSo 1 + 12 dy/dx - 2dy/dx + 12 dy/dx = 0\nThus 22dy/dx = -1\ndy/dx = -1/22\nie slope of tangent at (1, 6) is -1/22\n\n#3\nFind all open intervals on which the following function is decreasing:\nf(x) = x / (x^2 + x -2)\n= x/[(x + 2)(x - 1)]\n= 2/[3(x + 2)] + 1/[3(x - 1)] (by partial fractions)\nDiscontinuities at x = -2 and x = 1\n\nf'(x) = -2/[3(x + 2)²] - 1/[3(x - 1)²]\n\nSo f'(0) = -2/[3(2)²] - 1/[3(-1)²]\n= -2/12 - 1/3\n= -1/2 ***\n\nClearly f'(x) < 0 for all x except x = -2 and x = 1 where it is undefined \n\nie the function is decreasing fo x< -2, -2 < x < 1 and x > 1
135512
The first derivative of this is\n\n10x^4 - 32x^3 + 21x^2 - 20x + 10\n\nSet that equal to zero and solve for the FOUR solutions. That will give you the x values of the four turning points that are the maxima and minima of the original equation. \n\nThe second derivative will serve two purposes. It will enable to you to find the thtree inflection points, and also enable you to tell which of the four turning points of the curve are maxima and which of them are minima.\n\nThe second derivative is \n\n40x^3 - 96x^2 + 42x - 20\n\nSet that equal to zero and solve. The three solutions give you the x-values of the three inflection points. To use the second derivative to find which turning points are maxima and which are minima, plug in the value of x for each turning point into the second derivative. If the value of the second derivative is positive, the turning point in a minimum. If it is negative, the turning point is a maximum.
72955
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
399906
a) Use implicit differentiation.\n\n2x + 8y(dy/dx) = 0 + 3y + 3x(dy/dx)\n[remember to use the chain rule and the product rule]\n8y(dy/dx) - 3x(dy/dx) = 3y - 2x\n(dy/dx)(8y-3x) = 3y - 2x\ndy/dx = (3y-2x) / (8y-3x)\n\nb) A horizontal tangent means that dy/dx = 0. Use dy/dx from part (a).\n0 = (3y-2(3)) / (8y-3(3))\nNow solve for y... y = 2\nPoint P is (3,2)\n\nc) d2y/dx2 is the second derivative of the function. So do implicit differentiation on the derivative we already found. Remember to use the quotient rule.\nd2y/dx2 = [ (8y-3x)(3(dy/dx)-2) - (3y-2x)(8(dy/dx) -3) ] / (8y-3x)^2\nSince you have point P = (3,2) you don't have to simplify, just plug in.....\nd2y/dx2 = about -.2857, or -14/49\nA local maximum or minimum means that f' is 0, which they tell you in part (b). To find which, you have to see whether f' changes from positive to negative or from negative to positive. Since the slope of f', or f'' at point P is negative, it changed from positive to negative. So the point P is a maximum because the function was increasing until point P, then decreased after.\n\nI hope this helped!
399803
Welcome to calculus and its often confusing problems.  The reward comes when you finally realize how simple it all is!\n\nIn problem (a), the first two terms are straightforward, but the third term needs the "Chain Rule" for its derivative, because that term is built from "a function within a function."\n\nLet's focus on that term, and then do the others after:\n\n3e^(x^3) means "the coefficient is 3, which multiplies the exponential 'e to the power of x cubed'."  The 'x cubed' is the "function within another function" ... the so-called "other function" is simply the e^.\n\nChain Rule comes to the rescue here.  It says, basically, "do the outermost derivative and then work your way inside, multiplying the answer by each inside function's derivative, and so on."\n\nDo you know how to take the derivative of e^x?  Sure, it's just e^x (hint: it's also 1e^x, but we never write the 1 part).\n\nHow about e^(2x)?  Answer: 2e^(2x)\n\nSee how the "inside function" (2x) has a derivative (2) that appears as a multiplier?\n\nThat's going to happen with 3e^(x^3), too!  The derivative of "x cubed" is "three x squared", and Chain Rule tacks that on:\nDerivative of 3e^(x^3) is now 3 (3x^2) e^(x^3).\nSimplifying that a bit gives: 9 x^2 e^(x^3).\nThis is FAR uglier-looking now since I am unable to display exponents neatly here!\n\nThe rest of f(x)'s derivative should be straightforward for you:\n\nf'(x) = 4x^3 + 7 + 9(x^2)e^(x^3)\n\n(B)  To find g(x)'s derivative, you again use Chain Rule.  But it's helpful to review what cosine's derivative is (hint: it's -sin).  Just be aware, now, that the "inside function" is not just plain x (whose derivative is 1 and doesn't ever get written!) ... the inside function is now x^4.\n\nWhat is x^4's derivative?  It's 4x^3.  We'll be tacking that onto the front of our -sin stuff.\n\nFinally:\n\ng'(x) = -(4x^3)sin(x^4)\n\nThe "inside function" has been copied just where it was all along -- inside the cos or sin function.  But Chain Rule made it "spit out" its derivative in front (some teachers write it out following the cos or sin function, and then simplify by moving it in front in a later step).\n\nHope this helps!
196224
Take the derivative:\n\nf'(x) = 2x - 4\n\nSet the derivative equal to 0, and solve for x:\n\n0 = 2x - 4\nx = 2\n\nSub this value of x into the original equation to find the turning point:\n\nf(2) = (2)^2 - 4(2) + 1\nf(2) = -3\n\nSo, your turning point is (2, -3).\n\nAnd, since the x-squared term in the equation is positive, you know that this point is a minimum.\n\nDone and done.
490684
first, I would put it in an easier to express format g(t)=[t^(1/2)](1-t)  then take the derivative g'(t)=[(1/2)t^(-1/2)](1-t)-t^(1/2)   now just solve for g'(t)=0, so 0=[(1/2)t^(-1/2)](1-t)-t^(1/2)      now we distribute to get 0=[(1/2)t^(-1/2)]-(3/2)t^(1/2)     we add (3/2)t^(1/2) to both sides to get (3/2)t^(1/2)=(1/2)t^(-1/2)   simplfy this to [3sqrt(t)/2]=[1/2sqrt(t)]   we multiply both sides by 2sqrt(t) to get 6t/2=1 or 3t=1, which means that t=1/3\n\nAnd by the way, this isn't the chain rule, this is the Product Rule