d [ 2e^(-3x).cos[2x] ]/dx = d[2e^(-3x)]/dx .cos[2x] + 2e^(-3x).d[cos[2x]]/dx
...................................dy/dx = -6e^(-3x).cos[2x] - 4e^(-3x).sin[2x]
...................................dy/dx = -e^(-3x)(6cos[2x] + 4sin[2x])
...................................dy/dx = -2e^(-3x)(3cos[2x] + 2sin[2x])
sub (x=0) into dy/dx = -2e^(-3(0))(3cos[2(0)] + 2sin[2(0)])
............................= -6
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y = secx = 1/cosx
Result using implicit differentiation:
y = 1/cosx
ycosx = 1
dy/dx.cosx - ysinx = 0
dy/dx.cosx = ysinx
dy/dx = ysinx/cosx, Now sub in y = 1/cosx ...
dy/dx = sinx/cos²x
dy/dx = (sinx/cosx)(1/cosx) = tanxsecx
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y = cosecx = 1/sinx
Using Implicit Differentiation:
y = 1/sinx
ysinx = 1
dy/dx.sinx + ycosx = 0
dy/dx.sinx = - ycosx
dy/dx = - ycosx/sinx Now sub in y = 1/sinx ...
dy/dx = -(1/sinx)(cosx/sinx)
dy/dx = -cosec[x]cot[x]
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cosecx + secx = 1/sinx + 1/cosx
1/sinx + 1/cosx = (cosx + sinx) / sin[x]cos[x]
Now use identity 2sin[x]cos[x] = sin[2x]
sin[x]cos[x] = ½sin[2x] , sub this result into equation:
.... (cosx + sinx) / ½sin[2x]
.... 2(cosx + sinx) / sin[2x]