Does anyone have any idea how to do the Implicit Differentiation in Calculus?
965952
Implicit differentiation is really jsut an exercise in using the chain rule. For instance, if you had e^(x^2), its derivative would be e^(x^2) * 2x, since d(x^2)/dx = 2x. It's the same thing with implicit, except that you don't actually know what dy/dx is yet, so you just write dy/dx. So for my example, d(e^y)/dx = e^y dy/dx. Now, onto your problem:\n\nd(x/(x+y))/dx = ((x+y) - (1+dy/dx)x)/(x+y)², or being simplified, (y-x dy/dx)/(x+y)²\n\nOf course, this assumes you are differentiating with respect to x. If you are differentiating w.r.t. a third variable (e.g. t), then you would have to use the chain rule on x as well. So your derivative would be:\n\nd(x/(x+y))/dt = (dx/dt (x+y) - (dx/dt + dy/dt)x)/(x+y)²
766685
You need to differentiate the equation y=2x^2. \n\ndy/dx = 4x therfore the gradient at x=3 = 4*x = 12
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.
761317
We have to find derivative of, y = the integral where lower limit is e^x and upper limit is 0 of sin^3 t dt.\n\ny = int(e^x)(0) [sin^3 t] dt     [I would write lower limit in first bracket and upper limit in bracket after that, so dont confuse]\n\nNow, the derivative of the integrand f(t)=sin^3 t is f'(t)=3 (sin^2 t) *cos t\ndy /dt= d {int(e^x)(0) [sin^3 t] dt }/dt\n=>dy /dt= d {f(t)}/dt (at 0) - d {f(t)}/dt (at e^x)\n=>dy /dt= f'(t) (at 0) - f(t) (at e^x)\n=>dy /dt= [3 (sin^2 t) *cos t] (at 0) - [3 (sin^2 t) *cos t ] (at e^x)\n=>dy /dt= [3 (sin^2 0) *cos 0] - [3 (sin^2 (e^x)) *cos (e^x) ] \n=>dy /dt= [3 (0) *1] - [3 (sin^2 (e^x)) *cos (e^x) ] \n=>dy /dt= -[3 (sin^2 (e^x)) *cos (e^x) ]\n\nI hope that answers your question.\n\nAll the bset
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.
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....
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
255922
Im not sure if you need clarification on how to do this. Ill provide a few details on how to get these answers:\n\n1) ∫sinxdx = - cos(x) + C\n-This one is relatively straightforward, just think "what gives me sin(x) when I take its derivative"\n\n2)∫e3xdx = (Im assuming this means e^(3x))\n         = (e^(3x))/3 + C\n-Just think of the chain rule backwards. If you take the derivative of e^3x you'll have to multiply by 3, so when you take the derivative divide by 3.\n\n3)∫cos3xdx = (Im assuming this is (cos^3)x)\n           = ∫((cos^2)x )(cos(x))\n           = ∫(1-(sin^2)x)((cos(x))\n      Now let u=sin(x)\n             du=cos(x)dx\n           = ∫(1-u^2)du\n           = [u - (u^3)/3]\n     Sub back in for u\n           = sin(x) - (sin^3(x))/3 + C\n-Wow, that one was hard to type out. I hope you can follow it somewhat. The trick is to write cos^3(x) as cos^2(x) and cos(x). Then the identity cos^2(x) + sin^2(x)=1 is used. The u substution allows us to find the integral at the end.\n\n4) Derivative of 3x^4 + 2x^2 + x + 3\n                  = 12x^3 + 4x + 1\n-The derivative is defined by some long equation, but you can use the following fact:\nIf f(x)=x^n\nthen f'(x)=nx^n-1\n\nHopefully you understand most of what I did. Im sure someone else will probably explain a lot of this in much better detail. Like I said, I wasnt sure how much detail you were looking for. \n\nPS Ill tell you site that will help you forever in any integral calculus course: http://integrals.wolfram.com/index.jsp
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
274298
Use implicit differentiation to find dy/dx. Standard differentiation rules apply, except when you differentiate a term with a "y" in it you must tack on a dy/dx, so:\n\n2x+8y(dy/dx)=0\ndy/dx = -x/4y\n\nThe tangent line intersects the ellipse at (x1,y1), so the slope of the tangent line will be -x1/(4*y1).\n\nWe can also compute the slope of the tangent line by plugging the two points (-5,0) and (x1,y1) into the standard slope formula m = delta(y)/delta(x) and get that the slope of the tangent line is y1/(x1+5). Since these expressions both describe the slope of the tangent line, they must be equal, so\n-x1/(4*y1)=y1/(x1+5)\nx1^2+5*x1=-4*y1^2\nx1^2+4*y1^2+5*x1=0\n\nSince (x1,y1) is a point on the ellipse, it must satisfy the equation of the ellipse, so x1^2+4*y1^2=5. Substituting this in to x1^2+4*y1^2+5*x1=0 gives us 5+5*x1=0, and x1=-1. At this point, there are actually two y1 values that will satisfy the equation of the ellipse, y1 could be either plus or minus 1 (get these by letting x=-1 in the original equation of the ellipse), and so there are actually two points on the ellipse where the tangent line passes through (-5,0), and those points are (-1,1) and (-1,-1).
752772
You need to use the chain rule a couple of times.\nHow do you differentiate (something)^2 ? Make it 2*something. Then multiply by (something) differentiated.\nIn this case, (sin 3x)^2 differentiates to 2(sin 3x) * (sin 3x differentiated).\nNow, how do you differentiate sin(something)? Make it cos(something), then multiply by (something) differentiated.\nSo sin 3x differentiates to cos 3x * 3.\nOverall, you get 2 sin 3x cos 3x * 3 = 6 sin 3x cos 3x.
38734
(2/3)x^3 + (5/2)x^2\nyou just take the antiderivitive of each term.  so you backtrack. instead of taking away one from the power, you add one and then divide the whole term by the new power. so the first term would be to the third power. now if you took the derivitive you would multiply by that power of three, so we need to cancel that out by dividing by three. so it's (2/3)x^3  because when you take the derivitive and that three comes down, it will be (3)(2)/(3) and the 3's will cancel out giving you 2x^2 which is what you started with. so the antiderivitive is (2/3)x^3.\n\nnow we do the second term. so we add a power, so it would be squared. now we have to cancel that power out so we'll divide the whole thing by two(or rather, multiply by a half) so the antiderivitive of 5x would be (5/2)x^2. this can be checked by finding the derivitive. multiply the (5/2) times 2 and you are left with 5 and take a power away to get 5x.  integrals are easier than you think, its just the opposite of differentiation. the general formula is:  X^n integrated is (X^n+1)/n+1 so if it's x^3 it would become x^4/4 or (1/4)x^4. again, you can check this by differentiating it. the exponent of 4 is multiplied and cancels out the fraction, and the exponent turns to a 3 and you're left with what you started with...\nhope you understand integrals now :)
899944
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
117843
Sure.  The derivative of x^3 is 3x^2.  That should get you started.
795130
y = tan(πx - 1)\nlet u = πx - 1\nnow, dy/dx = (dy/du)(du/dx)\nd(tan(u)/du = sec^2(u)\ndu/dx = π, so \ndy/dx = πsec^2(u)\ndy/dx = πsec^2(πx - 1)
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]
1016838
Use the quotient rule for differentiation.  Taking the case where the firetruck is approaching:\n\nThe numerator of the derivative has the denominator of the original function times the derivative of the numerator minus the numerator of the original function times the derivative of the denominator.  \n\nThis is pretty easy, since the numerator of the original function is a constant.  The derivative of a constant is 0.  When approaching, you only have to worry about the original numerator times the derivative of (331 - v).  The 331 is a constant with a derivative of 0 leaving you the derivative of -v.  That's -1.  \n\nTherefore, the numerator of your derivative is -132400 * -1, or 132400 when approaching.\n\nThe denominator of the derivative is just the square of the original denominator, or (331 - v)^2 when approaching.  When approaching, that would give you 301^2 (301 squared), or 90601. \n\nWhen approaching, the rate of change would be 132,400/(301^2), or about +1.46.\n\nFinding the rate of change when moving away works the same, but with a +v instead of a -v.