What is the d/(dθ)(4/(2cos(θ)-sin(θ))) ?

The d/(dθ)(4/(2cos(θ)-sin(θ))) is -(4(-2sin(θ)-cos(θ)))/((2cos(θ)-sin(θ))^2)

What is the first d/(dθ)(4/(2cos(θ)-sin(θ))) ?

The first d/(dθ)(4/(2cos(θ)-sin(θ))) is -(4(-2sin(θ)-cos(θ)))/((2cos(θ)-sin(θ))^2)

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

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

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the chain rule:

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

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

\frac{d}{d θ} (2 cos (θ) - sin (θ)) = - 2 sin (θ) - cos (θ)

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

Simplify

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

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

f (x) = \frac{8 x}{x ^{2} + 1}, (1, 4)

\frac{d}{d x} (\frac{4 x}{3 - tan ( x )})

x \to 0 lim (sin (a x))

\frac{10 s}{s ^{2} + 16}

\int \frac{1}{sin ^{2} ( 2 x + \frac{π}{4} )} d x

What is the d/(dθ)(4/(2cos(θ)-sin(θ))) ?
The d/(dθ)(4/(2cos(θ)-sin(θ))) is -(4(-2sin(θ)-cos(θ)))/((2cos(θ)-sin(θ))^2)
What is the first d/(dθ)(4/(2cos(θ)-sin(θ))) ?
The first d/(dθ)(4/(2cos(θ)-sin(θ))) is -(4(-2sin(θ)-cos(θ)))/((2cos(θ)-sin(θ))^2)