# dy by dx of math?

Find dy/dx of

(i) y=x/[(1+x)(1-2x)]

(ii) y=(1+sin x)/(1+ cos x)

(iii) y=sin 3x/(1-x)^3

### 1 Answer

- cidyahLv 73 years ago
i)

y = x / (1-2x+x-2x^2)

y = x /(1-2x^2-x)

u(x)= x

u'(x) = 1

v(x) = 1-2x^2-x

v'(x) = -4x-1

dy/dx = (v(x) u'(x) - u(x) v'(x)) /v^2(x) --- quotient rule

dy/dx = ( (1-2x^2-x) (1) - x (-4x-1) ) / (1-2x^2-x)^2

dy/dx = ( 1-2x^2-x+4x^2+x) /(1-2x^2-x)^2

dy/dx = (2x^2+1) /(1-2x^2-x)^2

ii)

y = (1+sin x) /(1+cos x)

u(x) = 1+sin x

u'(x) = cos x

v(x) = 1+ cos x

v'(x) = -sin x

dy/dx = (v(x) u'(x) - u(x) v'(x)) /v^2(x) --- quotient rule

dy/dx = ( (1+cos x) (cos x) - (1+sin x) (-sin x)) / (1+cos x)^2

dy/dx = ( cos x + cos^2 x + sin^2 x +sin x) / (1+cos x)^2

dy/dx = (cos x + sin x + 1) / (1+cos x)^2

iii)

y = sin 3x /(1-x)^3

u(x) = sin 3x

u'(x) = 3 cos 3x

v(x) = (1-x)^3

v'(x) = 3(1-x)^2 (-1) = -3(1-x)^2

dy/dx = (v(x) u'(x) - u(x) v'(x)) /v^2(x) --- quotient rule

dy/dx = ((1-x)^3 (3 cos 3x) - sin 3x (-3) (1-x)^2) /(1-x)^6

dy/dx = (3 cos 3x (1-x)^3 + 3 sin 3x (1-x)^2 ) / (1-x)^6

dy/dx = (1-x)^2 ( 3 cos 3x (1-x) + 3 sin 3x ) /(1-x)^6

dy/dx = ( 3 cos 3x + 3 sin 3x - 3x cos 3x) / (1-x)^4

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