Find dy/dx if y = root((x-3)(x^2 + 4)/(3x^2 + 4x + 5)
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y = under root[(x-3)(x2 + 4)/ (3x2+4x +5) ]take log both sideslog y =under root [log[(x-3) (x2+4)/ (3x2+4x +5)]→ log y = 1/2 [ log(x-3) +log (x2 +4)- ( log(3x2 +4x +5) ]differentiating both sides1/y.dy/dx = 1/2 [ 1/(x-3). d/dx(x-3)+1/ (x2 +4) .d/dx(x2 +4) -1/(3x2 +4x +5).d/dx(3x2 +4x +5)]→dy/dx=y/2[ 1/(x-3). (1-0)+1/ (x2 +4) .(2x+0) -1/(3x2 +4x +5).(6x +4 +0)]→dy/dx=y/2[ 1/(x-3). (1-0)+1/ (x2 +4) .(2x+0) -1/(3x2 +4x +5).
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