cant figure out what the derivative would be...?
953139
f'(x)=(4/3)x^(1/3) +(4/3)x^(-2/3)
786389
The derivative is [(x-4)- (x+1)]/(x-4)^2 = -5/(x-4)^2\n\nIf we use the definition of the derivative:\n\n\ndy/dx = Lim h -> 0 {f(x+h) -f(x)}/h\n= lim h-> 0 [(x+h+1)/(x+h-4) - (x+1)/(x-4)]/h\n                   (x+h+1)(x-4)-(x+1)(x+h-4)\n= lim x-> 0 ---------------------------------------\n                   (x+h-4)(x-4)h\n\n= lim h-> 0 -5h/(x+h-4)(x-4)h = -5/(x-4)^2
49770
There is no closed form for the nth derivative of this function.  I symbolically calculated and factored the first 50 derivatives on MathCAD, and there was not a pattern.  Sorry, but you're out of luck....
176854
You can graph it but probably that is not the answer your teacher is looking for. If you wanna do this by the good old method:\n\nf'(x)=3^x*ln(3)\n\nYou are looking for a point where the slope equals to 1 meaning 1 equals to your derivative.\n\n3^x*ln(3)=1\n\nNow use the calculator: x=-0.0856
954163
you started right with the quotient rule (BT-TB/B^2), but the deriv of Tanx - 1 = sec^2 x, the -1 goes away as the dervi of a constant is zero (on the other hand, the dervi of a coefficient is that coefficient).  That is the critical mistake I saw... and changing that will change your whole answer. Just scratch it out and try again!
830837
If ∫f(x)dx = y(x), then f(x) = d y(x) / dx;  So to get the answer, take the derivative of the expression x^2 - 2x + 1.  The value of the integral at limit 1 is  a constant and will not appear in the derivative.
754733
Let y = u/v   where u=some function and v =some funtion\n\nthen dy/dx = (v*du/dx - u*dv/dx)/ (v^2)     **Quotient Rule**\n\nIn this case u = 6x^2-1 and v = 2*sqrt(x)\n\nSo you just differentiate and plug into the equation for dy/dx
750296
-2x + h\nIn about 5 minutes you will be taken the limit of the expression as h ---> 0.  Then, you will be told that then it will be the derivative.\n\nAfter that you take 2 - x^2 do the derivative and get -2x.
842257
divide through\n\ny = [x^(3/2)]/3 - 2x^(1/2) -x^(-1/2)\n\nthen take the derivative\n\ndy/dx = [x^(1/2)]/2 - x^(-1/2) + [x^(-3/2)]/2\n\nthen put x^(3/2) in the denominator\n\ndy/dx = [1/x^(3/2)]*{ [x^2]/2 - x + 1/2 }\n\nwe take out the one half to find\n\ndy/dx = [1/2x^(3/2)] * [ x^2 - 2x + 1 ]\n\nwhich is the same as writing \n\ndy/dx = [1/2x^(3/2)] * (x - 1)^2\n\nor a = -1 and b = 2
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
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, ∞)
1166733
dy/dx = {(3x^2-2)(4x-3)^4 - (x^3-2x)(4)(4x-3)^3(4)}/(4x-3)^8\n= {(3x^2-2)(4x-3)^4 - 16(x^3-2x)(4x-3)^3}/(4x-3)^8\n\nUsing the quotient rule.  d/dx f(x)/g(x) = {f'(x)g(x) - g'(x)f(x)}/g(x)^2
831486
Differentiate both sides, we have \n\ndy =  (1/x)dx + (1 + tan^2(y))/tany dy\n\ntherefore\n(1 - (1 + tan^2(y))/tany)dy=(1/x)dx,\n\ni.e.\n\ndy/dx=tany/(x(tany - tan^2- 1)).
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
1138866
I'll lead you in the most important part, but it is up to you to complete it.\n\nDerivative of x . y = x'.y + x.y'\n\nThen, you will have:\n\nf'(x) = (2x . ln x^2) + x^2 (ln x^2)'\n\nContinue from here, the only problem that you may have is finding the derivative of ln y , which you can see in the link below.
372419
I fear you have a missing element here... do you want the differencial of 4P^2..etc because if so you need to know what 4P^2..etc equals, or is 4P^2..etc the differencial of another expression, because if so your question is a bit like saying what is the answer to x=2\n\nIf you want to differencial of 4P^2..etc then I suggest you solve the original equation for t which will make the algebra easier and then use the chain rule where dP/dt = [1/(dt/dP)]
986880
Do you want how to derive them or what the derivatives are?\n\nF(x)=a^x  ---> f'(x) = a^x (ln a)\n\nf(x) = sin x ---> f'(x) = cos x\n         cos x      --->           -sin x\n         tan x       ---->           sec^2 (x)\n         sec x      ---->           sec x tan x\n         csc x      ---->           - csc x cot x\n         cot x       ---->            - csc^2 (x)