What is the derivative of f(x)=(cos(x)-4)/(2sin(x)) ?

The derivative of f(x)=(cos(x)-4)/(2sin(x)) is (-sin^2(x)-cos(x)(cos(x)-4))/(2sin^2(x))

What is the first derivative of f(x)=(cos(x)-4)/(2sin(x)) ?

The first derivative of f(x)=(cos(x)-4)/(2sin(x)) is (-sin^2(x)-cos(x)(cos(x)-4))/(2sin^2(x))

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derivative of

f (x) = \frac{cos ( x ) - 4}{2 sin ( x )}

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

Solution steps

\frac{d}{d x} (\frac{cos ( x ) - 4}{2 sin ( x )})

Take the constant out:

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

= \frac{1}{2} \frac{d}{d x} (\frac{cos ( x ) - 4}{sin ( x )})

Apply the Quotient Rule:

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

= \frac{1}{2} \cdot \frac{\frac{d}{d x} ( cos ( x ) - 4 ) sin ( x ) - \frac{d}{d x} ( sin ( x ) ) ( cos ( x ) - 4 )}{( sin ( x ) ) ^{2}}

\frac{d}{d x} (cos (x) - 4) = - sin (x)

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

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

Simplify

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

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

\frac{1}{3} t^{3} + 7 t^{2} + 50 t

\frac{d}{d x} ((x^{4} + 3 x^{2}) (1 - 5 x))

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

\frac{\partial}{\partial x} (\frac{x ^{2} - 2 x}{( x - 1 ) ^{2}})

\int e^{- b x} d x

What is the derivative of f(x)=(cos(x)-4)/(2sin(x)) ?
The derivative of f(x)=(cos(x)-4)/(2sin(x)) is (-sin^2(x)-cos(x)(cos(x)-4))/(2sin^2(x))
What is the first derivative of f(x)=(cos(x)-4)/(2sin(x)) ?
The first derivative of f(x)=(cos(x)-4)/(2sin(x)) is (-sin^2(x)-cos(x)(cos(x)-4))/(2sin^2(x))