What is the d/(dy)((1-cos(y))/(1-cos(2y))) ?

The d/(dy)((1-cos(y))/(1-cos(2y))) is (sin^3(y)-sin(2y)(-cos(y)+1))/(2sin^4(y))

What is the first d/(dy)((1-cos(y))/(1-cos(2y))) ?

The first d/(dy)((1-cos(y))/(1-cos(2y))) is (sin^3(y)-sin(2y)(-cos(y)+1))/(2sin^4(y))

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\frac{d}{d y} (\frac{1 - cos ( y )}{1 - cos ( 2 y )})

\frac{sin ^{3} ( y ) - sin ( 2 y ) ( - cos ( y ) + 1 )}{2 sin ^{4} ( y )}

Solution steps

\frac{d}{d y} (\frac{1 - cos ( y )}{1 - cos ( 2 y )})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d}{d y} ( 1 - cos ( y ) ) ( 1 - cos ( 2 y ) ) - \frac{d}{d y} ( 1 - cos ( 2 y ) ) ( 1 - cos ( y ) )}{( 1 - cos ( 2 y ) ) ^{2}}

\frac{d}{d y} (1 - cos (y)) = sin (y)

\frac{d}{d y} (1 - cos (2 y)) = 2 sin (2 y)

= \frac{sin ( y ) ( 1 - cos ( 2 y ) ) - 2 sin ( 2 y ) ( 1 - cos ( y ) )}{( 1 - cos ( 2 y ) ) ^{2}}

Simplify

\frac{sin ( y ) ( 1 - cos ( 2 y ) ) - 2 sin ( 2 y ) ( 1 - cos ( y ) )}{( 1 - cos ( 2 y ) ) ^{2}} : \frac{sin ^{3} ( y ) - sin ( 2 y ) ( - cos ( y ) + 1 )}{2 sin ^{4} ( y )}

= \frac{sin ^{3} ( y ) - sin ( 2 y ) ( - cos ( y ) + 1 )}{2 sin ^{4} ( y )}

\frac{150 t}{t + 3}

\int t 9 t^{2} + 4 d t

x^{2}, x^{4} - 3 x^{2}, - 2, 2

x \to a lim (x^{s i n (x)})

\frac{x ^{2}}{2 x + 4}

What is the d/(dy)((1-cos(y))/(1-cos(2y))) ?
The d/(dy)((1-cos(y))/(1-cos(2y))) is (sin^3(y)-sin(2y)(-cos(y)+1))/(2sin^4(y))
What is the first d/(dy)((1-cos(y))/(1-cos(2y))) ?
The first d/(dy)((1-cos(y))/(1-cos(2y))) is (sin^3(y)-sin(2y)(-cos(y)+1))/(2sin^4(y))