What is the derivative of (1-sin(x)/(1+cos(2x))) ?

The derivative of (1-sin(x)/(1+cos(2x))) is (-cos^3(x)+sin(2x)(-sin(x)+1))/(2cos^4(x))

What is the first derivative of (1-sin(x)/(1+cos(2x))) ?

The first derivative of (1-sin(x)/(1+cos(2x))) is (-cos^3(x)+sin(2x)(-sin(x)+1))/(2cos^4(x))

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\frac{d}{d x} (\frac{1 - sin ( x )}{1 + cos ( 2 x )})

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

Solution steps

\frac{d}{d x} (\frac{1 - sin ( x )}{1 + cos ( 2 x )})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 1 - sin ( x ) ) ( 1 + cos ( 2 x ) ) - \frac{d}{d x} ( 1 + cos ( 2 x ) ) ( 1 - sin ( x ) )}{( 1 + cos ( 2 x ) ) ^{2}}

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

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

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

Simplify

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

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

f (x) = - 1

\frac{d}{d z} (\frac{x}{y} - y z)

x \to 2 - lim (cos (\frac{1}{x - 2}))

\frac{d}{d x} (x^{3} e^{- 8 x})

x \to 0 lim (4 x^{2} - 5 x - 1)

What is the derivative of (1-sin(x)/(1+cos(2x))) ?
The derivative of (1-sin(x)/(1+cos(2x))) is (-cos^3(x)+sin(2x)(-sin(x)+1))/(2cos^4(x))
What is the first derivative of (1-sin(x)/(1+cos(2x))) ?
The first derivative of (1-sin(x)/(1+cos(2x))) is (-cos^3(x)+sin(2x)(-sin(x)+1))/(2cos^4(x))