What is the derivative of (6cos(x)/(1-sin(x))) ?

The derivative of (6cos(x)/(1-sin(x))) is 6/(1-sin(x))

What is the first derivative of (6cos(x)/(1-sin(x))) ?

The first derivative of (6cos(x)/(1-sin(x))) is 6/(1-sin(x))

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\frac{d}{d x} (\frac{6 cos ( x )}{1 - sin ( x )})

\frac{6}{1 - sin ( x )}

Solution steps

\frac{d}{d x} (\frac{6 cos ( x )}{1 - sin ( x )})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 6 \frac{d}{d x} (\frac{cos ( x )}{1 - sin ( x )})

Apply the Quotient Rule:

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

= 6 \cdot \frac{\frac{d}{d x} ( cos ( x ) ) ( 1 - sin ( x ) ) - \frac{d}{d x} ( 1 - sin ( x ) ) cos ( x )}{( 1 - sin ( x ) ) ^{2}}

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

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

= 6 \cdot \frac{( - sin ( x ) ) ( 1 - sin ( x ) ) - ( - cos ( x ) ) cos ( x )}{( 1 - sin ( x ) ) ^{2}}

Simplify

6 \cdot \frac{( - sin ( x ) ) ( 1 - sin ( x ) ) - ( - cos ( x ) ) cos ( x )}{( 1 - sin ( x ) ) ^{2}} : \frac{6}{1 - sin ( x )}

= \frac{6}{1 - sin ( x )}

\int_{1}^{1} x^{2} d x

\int - e^{- x} cos (3 x) d x

\frac{d}{d x} (2 x^{3} e^{2 x})

\frac{\partial}{\partial y} (x^{2} e^{- yz})

\int \frac{1}{2 + 5 y ^{2}} d y

What is the derivative of (6cos(x)/(1-sin(x))) ?
The derivative of (6cos(x)/(1-sin(x))) is 6/(1-sin(x))
What is the first derivative of (6cos(x)/(1-sin(x))) ?
The first derivative of (6cos(x)/(1-sin(x))) is 6/(1-sin(x))