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

The derivative of (sin(x)/(1-cos(x))) is 1/(-1+cos(x))

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

The first derivative of (sin(x)/(1-cos(x))) is 1/(-1+cos(x))

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

\frac{1}{- 1 + cos ( x )}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

\frac{d}{d θ} (sin (θ))

x \to - 5 lim (\frac{( x - 5 )}{x + 5})

8 t^{4}

\int \frac{( x + 4 )}{x ^{2} + 8 x + 17} d x

y^{^{'}} + \frac{2}{t} y = \frac{cos ( t )}{t ^{2}}, y (π) = 0

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