Derivative of [ y = (tanx - 1) / secx)]?
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!
971568
What's the point of asking the same question twice?\n\nIf the issue is that you're only allowed to work on one side of the equation or the other, then you can pick one side, follow down to the bottom, and do everything in reverse to match the other side.\n\nRHS:\n\n1 + tan²x / tanx = [1 + (sec²x - 1)] / tanx = sec²x / tanx\n= 1/cos²x / (sinx / cosx) = 1/cos²x * cosx/sinx\n= 1/(sinx cosx) = cscx secx\n\nLHS:\n\n[cscx + secx]/[sinx + cosx] = [1/sinx + 1/cosx]/[sinx + cosx]\n= [(cosx + sinx)/(sinx cosx)] * 1/[sinx + cosx]\n= 1/(sinx cosx) = cscx secx
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
857727
According to the chain rule and to the rule for the derivative of a product, we have y' = (x^2 +3) e^(x^2+1) (2x) + (2x) e^(x^2 +1) = e^(x^2 +1) (2x^3 + 6x + 2x) = e^(x^2 +1) (2x^3 + 8x)
120626
There's a derivatives tutorial on the following page:\n\nhttp://www.karlscalculus.org/calculus.html#s4
350899
coshx\nand the derivative of coshx is sinhx
816542
Find derivatives with respect to each variable.
904114
First, put your original equation as Y1\n\nFor Y2, type  nDeriv(Y1, x, x)\n\npress the buttons   (math)(8)   to get   nDeriv\nthe buttons (VARS)(1)(1)  to get Y1\n\n\nFor Y3, type  nDeriv(Y2, x, x)\n\nthen just display graph Y3\n\nIt will take quite a long while to graph, better if you manually differentiate the original equation once, then use the nDeriv to do second derivative.
837557
(secx - tanx)^2 = (1/cosx - sinx/cosx)^2 = ((1-sinx)/cosx)^2\n\n(1-sinx)/(1+sinx) \n= [(1-sinx)(1-sinx)/(1+sinx)(1-sinx)]\n\n= [(1-sinx)^2/(1-(sinx)^2)] \n= [(1-sinx)^2/(cosx)^2] \n= ((1-sinx)/cosx)^2
927315
(tanx- cotx) / (tanx+cotx) , simplify by using cotx = 1/tanx\ntanx -1/tanx)/ (tanx +1/tanx)\n=(tan^2x -1)/ (tan^2x +1)\n=(sec^2x -2)/ sec^2x ........\n[ since sec^2x = 1+ tan^2x,    so sec^2x -2 = tan^2x - 1]  :)\n=1- ( 2/sec^2x)\n= 1- 2 cos^2x\nso, (tanx- cotx) / (tanx+cotx) + 2 cos^2x =\n                     1- 2 cos^2x +2 cos^2x = 1......  :D
765615
differentiate implicitly:\n\n2x + 2yy' = 0\n\nand then some algebra\n\ny' = -(x/y)\n\nthen differentiate again:\n\ny'' = (-y - (-x)*y')/y^2 = (-y + xy')/y^2\n\nthen substitute in values:\n\ny' = -0/5 = 0\n\nand y'' = (-5 + 0*0)/(5^2) = =-5/25 = -1/5
850884
using chain rule and product rule\n\ny'= 3x^(2)*tan(4x)\n         +4x^(3)*sec^(2)(4x)\n  = x^2(3tan(4x)+4x*sec^(2)(4x) \n\ni hope that this helps
982407
dy/dx=cosx*e^y*e^sinx\ndy/e^y= cosx*e^sinx*dx\ne^-ydy= cosx*e^sinx*dx\nintegrate\ne^-y/(-1)=int. cosx*e^sinx*dx\nput sinx=t\ncosxdx=dt\n-e^-y=int.e^tdt\n        =e^t+c\n        =e^sinx+c\nput x=0,y=0\n-1=1+c\nc=-2\n-e^-y=e^sinx-2
888343
Use the chain rule:  dy/dx = d/dx[(ln x)^x] = x(ln x)^(x-1)d/dx[ln x]
408737
Need to use the product rule here:\n\nd/dx(1/2*tanxsinx)\n\n= [d/dx(1/2tanx)]sinx + 1/2tanx*d/dx(sinx)\n= (1/2sec^2x)sinx + 1/2tanx*cosx\n= sinx/2cos^2x + sinx/2cos^2x\n= tanx/cosx
713982
use logarthmic differentiation\n\ny = x^(1/x)\n\nln(y) = (1/x)* ln(x)\n\ndiff w.r.t x\n\n(1/y) (dy/dx) = (1/x)*(1/x) +ln(x)*(-1/x^2)\n\n(dy/dx) = { y/(x^2) }  [ 1 - ln(x) ]\n\nsubstitute for y
1162151
Use the fact that tan^2(x) = sec^2(x ) - 1 and integrate that.  You will get tanx -x +c.