What is the derivative of 8/(sin(x)+1/(cot(x))) ?

The derivative of 8/(sin(x)+1/(cot(x))) is (-8cot(x)+sec(x)tan(x))/(sin(x))

What is the first derivative of 8/(sin(x)+1/(cot(x))) ?

The first derivative of 8/(sin(x)+1/(cot(x))) is (-8cot(x)+sec(x)tan(x))/(sin(x))

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\frac{d}{d x} (\frac{8}{sin ( x )} + \frac{1}{cot ( x )})

\frac{- 8 cot ( x ) + sec ( x ) tan ( x )}{sin ( x )}

Solution steps

\frac{d}{d x} (\frac{8}{sin ( x )} + \frac{1}{cot ( x )})

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (\frac{8}{sin ( x )}) + \frac{d}{d x} (\frac{1}{cot ( x )})

\frac{d}{d x} (\frac{8}{sin ( x )}) = - \frac{8 cot ( x )}{sin ( x )}

\frac{d}{d x} (\frac{1}{cot ( x )}) = sec^{2} (x)

= - \frac{8 cot ( x )}{sin ( x )} + sec^{2} (x)

Simplify

- \frac{8 cot ( x )}{sin ( x )} + sec^{2} (x) : \frac{- 8 cot ( x ) + sec ( x ) tan ( x )}{sin ( x )}

= \frac{- 8 cot ( x ) + sec ( x ) tan ( x )}{sin ( x )}

\int_{0}^{\frac{π}{2}} cos^{2} (2 x) d x

sin^{4} (\frac{x}{4}) + cos^{4} (\frac{x}{4})

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

f (θ) = θ cos (θ) sin (θ)

\frac{\partial}{\partial y} (y^{2} - 4 y + x y)

What is the derivative of 8/(sin(x)+1/(cot(x))) ?
The derivative of 8/(sin(x)+1/(cot(x))) is (-8cot(x)+sec(x)tan(x))/(sin(x))
What is the first derivative of 8/(sin(x)+1/(cot(x))) ?
The first derivative of 8/(sin(x)+1/(cot(x))) is (-8cot(x)+sec(x)tan(x))/(sin(x))