¿what is the relationship between partial and total derivative?
813180
three variables are related to time ...( X,Y,Z,) related to time\n\nthe total derative can be expressed by the sum of the partials... that is dx/dt + dy/dt + dz/dt = total derative....\n\nthe partial with respect to x (dx/dt)  plus\nthe partial with respect to y (dy/dt)  plus\nthe partial with respect to z (dz/dt)  equals\n\nthe total derative\n\nthe partial gives us an idea of how the function acts on that varaible
1015799
Use the second derivative test, not sure exactly what it is still but it should be easy to find on the internet.
794869
Here's a good website that discuss implicit differentiation with some example problems.\n\nHope this helps
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)
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.
858220
Depends on what variable you're differentiating with respect to:\ndV/dr : 2/3*pi*r*h\ndV/dh:  1/3*pi*r^2
888343
Use the chain rule:  dy/dx = d/dx[(ln x)^x] = x(ln x)^(x-1)d/dx[ln x]
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!
114706
Change of variable t=Tan[x]\nwe have dx=(1+Tan[x]^2)dt\n\nSubstitute:\nIntegral[(1+Tan[x]^2)\n/(1-Tan[x]^2)]dx\n=\nIntegral[dt/(1-t^2)]\n=...\n\nusing partial fraction decomposiotion\n2/(1-t^2)=1/(t+1)-1/(t-1)\n\n...=\n(1/2)(Integral[dt/(t+1)]\n-Integral[dt/(t-1)])\n=\n(1/2)(Ln[Abs[t+1]]\n-Ln[Abs[t-1]])\n=\n(1/2)Ln[Abs[(t+1)/(t-1)]]\n\nreplacing back t we have\n\n(1/2)Ln[Abs[(Tan[t]+1)\n/(Tan[t]-1)]]\nplus C\n\nNote that Abs is needed!\nIf it is used then we can write the result as you requested:\n\n(1/2)Ln[Abs[(1+Tan[t])\n/(1-Tan[t])]]\nplus C
1177010
If the you have a local max/min at an interior point, , the partial derivatives must be zero.\nSince the sum of the second partials is zero, the point must be a saddle point, not a max/min.
972569
By definition of the derivative\ng'(x) = lim h=>0 [g(x + h) - g(x)] /h =  lim h=>0 [3xh + h^2]/h = \nlim h=>0 [3x + h] = 3x
671983
A derivative is the primary mathematical tool that is used to calculate and study rates of change.\n\nf'(Xo)=lim f(x+h)-f(x) / f  (/is divide which I am sure u know)\n          \n\nK so thats really confusing-not easy trying to write out the formula on this but hope it helps a little.
108610
If look to the integrator more deeply, one finds that 1/x is the derivative of ln(x).\nThen this question comes up: \nWhat is the integration of uu'dx?\n\nObviously the answer is u^2/2.\n\nSo our answer is (ln(x))^2/4+c.
1298
derivative of 5(sec(x)^2) with respect to sec(x) = 10(sec(x))\n\nderivative of 5(sec(x)^2) with respect to x = 10 * derivative of sec(x) with respect to x
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
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.
914813
Your correct, it is equal to f(x).\n\nthe integral and the differential cancel eachother out.!