what is the derivitive of ln x?
605811
As Lauren said, it's 1/x.\n\nMemorize this. It will be very useful.
723369
ln e^(3x) = ln4\n3x(lne) = ln4  (note lne = 1)\n3x = ln4\nx = (ln4)/3\nx=0.462
746085
taking log on both sides\nlny=(x-5)ln2\n1/y(dy/dx)=ln2\ndy/dx=yln2=2^(x-5)ln2
34808
F'(x) is the derivitive of F(x), which means it describes the change in F with respect to x.  Another way to say this is that F'(x) is the slope of F(x).  \n\nTo sketch the graph of F'(x), look at your graph of F(x).  Pick a point on that graph (maybe the first point). See what the slope (remember rise over run?) of F(x) is at that point.  That slope is the derivitive F' for that value of x.  Plot this on your F'(x) skecth.  Go the the next point and repeat.
865054
Let u=ln t, e^u=t, du=dt/t, dt=t du\nThen:\n∫ln² t dt = ∫t u² du = ∫u² e^u du\nIntegrating by parts:\nu² e^u - ∫2u e^u du\nu² e^u - 2u e^u + ∫2e^u du\nu² e^u - 2u e^u + 2e^u + C\nt ln² t - 2t ln t + 2t +C\nAnd we are done.
1060074
d (ln(x)) / dx = 1/x\n\nThe derivative of the natural log of x with respect to x equals 1 / x
898070
ln(x) + 2ln(4) = ln(7) - ln(x)\nln(x) + ln(4^2) = ln(7) - ln(x)\nln(x*4^2) = ln(7/x)\nln(16x) = ln(7/x)\n16x = 7/x\n16x^2 = 7\nx^2 = 7/16\nx = sqrt(7/16)\nx = sqrt(7)/4
882357
The derivative of ln|f(x)| is 1/f(x) * d f(x) / dx;  this will be\n\n1/[sec5x + tan5x] * d(sec5x) / dx + d(tan5x) / dx\n\nd(sec5x) / dx = d(1/cos5x) / dx = -1/cos^2(5x) * -5sin5x = 5tanx*secx\n\nd(tan5x) / dx = 5sec^2(5x)\n\n(5tanxsecx + 5sec^2(5x)) / (sec5x + tan5x)\n\nFactor out 5sec5x to get 5sec5x * (tan5x + sec5x)/(sec5x + tan5x) = 5sec5x
892157
wait 1 sec.\nOk this is reasonably difficult, you have to use the chain rule and the quotient rule.\nUsing the chain rule:\nderivative = [ (1/2) * ( (2x + 3) / (7x+4) ) ^ (-1/2) ]  *  [d/dx (2x+3) / (7x+4) ]\n\nSo now you to use the quotient rule to find:\nd/dx (2x+3) / (7x+4) = (14x+8 - 14x + 21) / (7x+4)^2\n= 29 / (7x+4)^2\n\nSo the derivative is [1/2 * ((2x+3)/(7x+4))^(-1/2)]*[29 / (7x+4)^2]\n\n\nBy the way i got the thumbs down before I wrote all this up, so it doesn't mean it's wrong.
775564
if ln(x)=y, then by the definition of ln, x=e^y\ne^(ln(x)) for ln(x)=y is e^y\nso e^ln(x)=e^y which we already showed =x which is what we wanted to show. Hope this helps. If not, email me.
798496
Let y = lnu where u = lnv where v = ln(6x)\n\ndy/dx = dy/du . du/dv . dv/dx\n= 1/u . 1/v . 6/6x\n= 1/ln(ln(6x)) . 1/ln(6x) . 1/x\n= 1/[xln(6x)ln(ln(6x))]
805634
It is easier (by far) to do this with logarithmic differentiation:\ndy/dx=y*d/dx[ln y].\n\nHere, with y equal to your expression,\nln y=ln 10 +ln x +4ln(x-1)\n-3ln(x-2)-2ln(x+1),\nso the derivative of the logarithm is\n1/x +4/(x-1) -3/(x-2)-2/(x+1).\nThe first derivative is now your original expression times this.\n\nTo find the second derivative, use the product rule. The derivative of y is already known, and the second derivative of the logarithm is easy.
773418
derivative of e^u = e^u du\nderivative of ln(u) = du/u\na^u=e^{uln(a)}, so the derivative of a^u =  e^{uln(a)} ln(a) = ln(a) a^u\ns
697108
e^(1+lnx) = e^1 * e^lnx = e*x = ex\nder(ex) = e+c
761104
The first derivative is correct. The second derivative is:\n\n- [ln 6x+1] / [ (xln(6x))^2 * ln(ln(6x))] - 1 / [ln(ln(6x)) * (xln(6x)) ]^2\n\nEssentially you want to use the product rule considering the first derivative to be:\n\n1/xln(6x) * 1/ln(ln(6x))\n\nd[1/xln(6x)]/dx = -1 * d[xln(6x)]/dx * [xln(6x)]^(-2)\n                        = -1 * [ln(6x)+1] / [xln(6x)^2]\n\nd[1/ln(ln(6x))]/dx = -1 * d[ln(ln(6x))]/dx * [ln(ln(6x))]^(-2)\n                           = -1 * [1 / xln(6x)] / [ln(ln(6x))]^2\n\nAnd then you use the product rule: (u * v)' = u' * v + u * v' to end up with what I gave you above.
350899
coshx\nand the derivative of coshx is sinhx
882612
2x^2 ln(x) - x^2\n\nVerify by differentiating\n2x + 4x ln(x) - 2x = 4x ln(x)