Question:
if y= under square root of (secx-tanx divided by secx tanx) then dy by dx=
Answer:
Given y = √(secx - tanx)/(secx*tanx)
=√{1/cosx - sinx/cosx}/{(1/cosx)*tanx} (secx = 1/cosx, tanx = sinx/cosx)
= √{(1-sinx)/cosx}/{tanx/cosx}
=> y = √{(1-sinx)/tanx}
Now take square on both side, we get
y2 = (1-sinx)/tanx
Differentiate w.r.t. x, we get
2y * dy/dx = {tanx*(-cosx) - (1-sinx)*sex2 x}/tan2 x
= {-tanx*cosx - (1-sinx)/cos2 x}/tan2 x
= [{-tanx*cos2 x - (1-sinx)}/cos2 x]/tan2 x
=[{-tanx*cos2 x - (1-sinx)}/cos2 x]/(sin2 x/cos2 x)
= {-tanx*cos2 x - 1 + sinx)}/sin2 x
= {-(sinx/cosx)*cos2 x - 1 + sinx}/sin2 x
= {-sinx*cosx - 1 + sinx}/sin2 x
= { sinx -sinx*cosx - 1}/sin2 x
=> dy/dx = (1/2y) *{ sinx -sinx*cosx - 1}/sin2 x
= (1/2)* [√{tanx/(1-sinx)}]*(sinx -sinx*cosx - 1)/sin2 x }
= (1/2)* [√{sinx/(1-sinx)*cosx}]*(sinx -sinx*cosx - 1)/sin2 x }
= (1/2)* [√{sinx/(1-sinx)*cosx}/sinx]*(sinx -sinx*cosx - 1)
=>dy/dx = (1/2)* [√1/{(1-sinx)*cosx}]*(sinx -sinx*cosx - 1)
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