What is the d/(dc)((sin(2c))/(2-cos(2c))) ?

The d/(dc)((sin(2c))/(2-cos(2c))) is (2cos(2c)(2-cos(2c))-2sin^2(2c))/((2-cos(2c))^2)

What is the first d/(dc)((sin(2c))/(2-cos(2c))) ?

The first d/(dc)((sin(2c))/(2-cos(2c))) is (2cos(2c)(2-cos(2c))-2sin^2(2c))/((2-cos(2c))^2)

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\frac{d}{d c} (\frac{sin ( 2 c )}{2 - cos ( 2 c )})

\frac{2 cos ( 2 c ) ( 2 - cos ( 2 c ) ) - 2 sin ^{2} ( 2 c )}{( 2 - cos ( 2 c ) ) ^{2}}

Solution steps

\frac{d}{d c} (\frac{sin ( 2 c )}{2 - cos ( 2 c )})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d c} ( sin ( 2 c ) ) ( 2 - cos ( 2 c ) ) - \frac{d}{d c} ( 2 - cos ( 2 c ) ) sin ( 2 c )}{( 2 - cos ( 2 c ) ) ^{2}}

\frac{d}{d c} (sin (2 c)) = cos (2 c) \cdot 2

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

= \frac{cos ( 2 c ) \cdot 2 ( 2 - cos ( 2 c ) ) - 2 sin ( 2 c ) sin ( 2 c )}{( 2 - cos ( 2 c ) ) ^{2}}

2 sin (2 c) sin (2 c) = 2 sin^{2} (2 c)

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

\int \frac{3}{1 - 9 x ^{2}} d x

\int (x^{2} - 3 x - 4) d x

f (x) = x + 4, (12, 4)

4 y^{^{''}} + 6 y^{^{'}} - 4 y = 0

\int 1 - cos (x) d x

What is the d/(dc)((sin(2c))/(2-cos(2c))) ?
The d/(dc)((sin(2c))/(2-cos(2c))) is (2cos(2c)(2-cos(2c))-2sin^2(2c))/((2-cos(2c))^2)
What is the first d/(dc)((sin(2c))/(2-cos(2c))) ?
The first d/(dc)((sin(2c))/(2-cos(2c))) is (2cos(2c)(2-cos(2c))-2sin^2(2c))/((2-cos(2c))^2)