Help me to find the Derivatives of this function including intergration?
452294
Okay, this is how I read this problem:\n\nThe integral of 1/(t-1) dt on the interval from 0 to (1-3x+x^2).\n\nFirst look at what you're integrating 1/(t-1) dt.  This can be rewritten (t-1)^-1 dt.  If you substitute u for t-1 and du for dt (the derivative of t-1 = 1) then you have the integral of u^-1 du.  This is ln |u| (ln is the natural log).  Sub t-1 for u and you get ln |t-1|.\n\nNow you use your interval and you get:\n\nln |(1-3x+x^2) - 1| - ln |0|\n\nSimplify and you get:\n\nln |x^2 - 3x| - ln |0|
754435
I | AUG is not correct.  You cannot differentiate power functions by the exponent rule.  (Even doing that, she made an error.)  The implicit differentiation done by Greg G is correct, or you can use the functional power rule:\n\nd(ƒ^g)/dx = (ƒ^g)*[(dƒ/dx)*(g/ƒ) + (dg/dx)*lnƒ]
859250
integral = x*tan(x)+ln(sec(x))  (use integration by parts with du = sec^2(x) and v = x).\n\ng'(x) = f(x), so a, c, e and g are true (b is true if f is continuous)\n\nSteve
853051
We have to find derivative of, g(x) = the integral (with lower limit tanx and upper limit x^2) of 1 / [sqrt(2+t^4)] dt.\n\nNow, the derivative of the integrand f(t)=1 / [sqrt(2+t^4)] is \nf'(t)= -2*t^3 / (2+t^4)^(3/2) [By Quotient Rule]\n\nd(g(x)) /dt= d {int(tan x)(x^2) {1 / [sqrt(2+t^4)] }dt} /dt\n=>dy /dt= d {f(t)}/dt (at tan x) - d {f(t)}/dt (at x^2)\n=>dy /dt= f'(t) (at tan x) - f(t) (at x^2)\n=>dy /dt= [2*t^3 / (2+t^4)^(3/2)] (at tan x) - [2*t^3 / (2+t^4)^(3/2)] (at x^2)\n=>dy /dt= [2*(tan x) ^3 / (2+(tan x)^4)^(3/2)]  - [2*(x^2)^3 / (2+(x^2)^4)^(3/2)] \n=>dy /dt= [2*(tan x) ^3 / (2+(tan x)^4)^(3/2)]  - [2*(x^8) / (2+(x^16))^(3/2)]\n\nI hope that answers your question.
909269
Take the derivative of the original function. For your given example, if I remember my calculus correctly:\nf(x)=y\nf'(x)=1\nGraph the new equation.
1159338
Integrate using the method of substitution based on the fact that the derivative of log(y) + C = 1/y\n\nlet u = log(y) then du/dy = 1/y. So du = (1/y) dy\n\nUsing these substitutions we have:\n\nInt 1/y*log(y) dy  = Int 1/log(y)*(1/y) dy\n\n= Int 1/u du  = log(u)+C = log(logy)+C, since u = log(y)\n\nThis is correct since from the chain rule:\n\n d/dy log(logy)+C= 1/log(y)*d(logy)/dy= 1/(y*logy)
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
919338
Well, I see you've used the correct order of operations this time. That's good - it shows that you're improving.\n\n#1: Apply the product and chain rules to find f'(x), then evaluate at 0:\n\nf'(x) = d(√(x³+5x+121))/dx (x²+x+11) + (2x+1)√(x³+5x+121)\n\nf'(x) = 1/(2√(x³+5x+121)) * d(x³+5x+121)/dx * (x²+x+11) + (2x+1)√(x³+5x+121)\n\nf'(x) = 1/(2√(x³+5x+121)) * (3x²+5) * (x²+x+11) + (2x+1)√(x³+5x+121)\n\nNow evaluating :\n\nf'(0) = 1/(2√121)*5*11 + √121 = 5/2 + 11 = 27/2\n\n#2: First, evaluate the function at y=1 to get the value of x:\n\nxy^2 -2y + 4y^3 = 6\nx - 1 + 4 = 6\nx = 3\n\nNow, implicitly differentiate both sides:\n\nd(xy^2 -2y + 4y^3)/dx = d(6)/dx\ny² + 2xy dy/dx - 2 dy/dx + 12y² dy/dx = 0\n\nSolve for dy/dx:\n\ny² + (2xy - 2 + 12y²) dy/dx = 0\n(2xy - 2 + 12y²) dy/dx = -y²\ndy/dx = -y²/(2xy - 2 + 12y²)\n\nNow evaluate at (3, 1):\n\ndy/dx = -1/(6 - 2 + 12) = -1/16\n\n#3: First, find the derivative:\n\nf'(x) = (x²+x-2 - (2x+1)x)/(x²+x-2)² (using the quotient rule)\n\nSimplifying:\n\nf'(x) = (x²+x-2 - 2x² - x)/(x²+x-2)²\nf'(x) = (-x²-2)/(x²+x-2)²\n\nNow we must find where f'(x) < 0\n\n(-x²-2)/(x²+x-2)² < 0\n-x²-2 < 0 (we know (x²+x-2)² is nonnegative, and if the expression on the left is defined, it is also nonzero. Thus it is positive and we can multiply by it without changing the inequality)\n-x² < 2\n\nNow, since x is a real number, this expression will always be true. Thus, if f'(x) is even defined, x is decreasing. Now f'(x) is defined provied that (x²+x-2)²≠0. So:\n\n(x²+x-2)²≠0\n(x-1)²(x+2)²≠0\nx≠1 and x≠-2\n\nWhich makes sense, since those are vertical asymptotes of the original function. Thus, the function is decreasing on:\n\n(-∞, -2) ∪ (-2, 1) ∪ (1, ∞)
888343
Use the chain rule:  dy/dx = d/dx[(ln x)^x] = x(ln x)^(x-1)d/dx[ln x]
1020565
The derivative of the function given is\n\ny'=cos(2x)*2 - 2*cos(x) = 2(2cos^2(x)-1)-2*cos(x) \n= 4cos^2(x)-2cos(x)-2 = 2(2cos^2(x)-cos(x)-1)\n=2(2cos(x)+1)(cos(x)-1)\n\nFor y'=0, either 2cos(x)+1=0 or cos(x)-1=0, from which we get\n\ncos(x)=-1/2 (which gives either x=2*pi/3 + 2*pi*k, k an integer or x=-2*pi/3 + 2*pi*k, k an integer)\n\nor \ncos(x)=1, so x=2*pi*k, k an integer.
846527
You'll have to do this in more than one step.  Basically it will sort of have to be a program.  I would start with the restart; command and then  define the function (I would look this up in the help menu since it is dependent on 2 variables).  Then use a command to find the derivative.  Then a command to solve for the pairs (x,y) when the derivative is 0.  That will give you 1.  For 2 and 3 you will also have to do a series of commands.  Use your calc book to refresh your memory on the procedure of finding these, it should be a similar procedure in maple.
844269
assuming that your question is V(w)=w^2*(3/(8w^3)),\nwe can simplify it as V(w)=3w^2/8w^3 = 3/8w\nso, the derivative will be V'(w) = -1*3/8w^2 = -3/8w^2
200344
W=f.delta(x) \nx=w/f = 49821/717= 69.49 metes
873172
u can get them with the help of second and higher derivatives.\ninflection point is there where the second derivative is zero.
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
1032260
1. if x=f(t) and y=f(t) then\n\nx^4+y^4=6y\n\nd/dt { x^4+y^4=6y}\n\n4x^3 (dx/dt) + 4y^3(dy/dt) =6(dy/dt)\n\nand you have to have the x=f(t) & y=f(t) in order to substitute for the derivatives.\n\n2. if y=f(x) and x=f(t) then\n\nx^4+y^4=6y\n\nd/dt { x^4+y^4=6y}\n\n4x^3 (dx/dt) + 4y^3(dy/dx)(dx/dt) =6(dy/dt)(dx/dt)\n\nthen substitute for the derivatives of the functions y=f(x) & x=f(t)