What is the d/(dθ)(5/(sin(θ)-2cos(θ))) ?

The d/(dθ)(5/(sin(θ)-2cos(θ))) is -(5(cos(θ)+2sin(θ)))/((sin(θ)-2cos(θ))^2)

What is the first d/(dθ)(5/(sin(θ)-2cos(θ))) ?

The first d/(dθ)(5/(sin(θ)-2cos(θ))) is -(5(cos(θ)+2sin(θ)))/((sin(θ)-2cos(θ))^2)

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\frac{d}{d θ} (\frac{5}{sin ( θ ) - 2 cos ( θ )})

- \frac{5 ( cos ( θ ) + 2 sin ( θ ) )}{( sin ( θ ) - 2 cos ( θ ) ) ^{2}}

Solution steps

\frac{d}{d θ} (\frac{5}{sin ( θ ) - 2 cos ( θ )})

Take the constant out:

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

= 5 \frac{d}{d θ} (\frac{1}{sin ( θ ) - 2 cos ( θ )})

Apply exponent rule:

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

= 5 \frac{d}{d θ} ((sin (θ) - 2 cos (θ))^{- 1})

Apply the chain rule:

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

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

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

= 5 (- \frac{1}{( sin ( θ ) - 2 cos ( θ ) ) ^{2}} (cos (θ) + 2 sin (θ)))

Simplify

5 (- \frac{1}{( sin ( θ ) - 2 cos ( θ ) ) ^{2}} (cos (θ) + 2 sin (θ))) : - \frac{5 ( cos ( θ ) + 2 sin ( θ ) )}{( sin ( θ ) - 2 cos ( θ ) ) ^{2}}

= - \frac{5 ( cos ( θ ) + 2 sin ( θ ) )}{( sin ( θ ) - 2 cos ( θ ) ) ^{2}}

x^{2} \frac{d y}{d x} = x - x y

y^{^{'}} = 9 x + y, y (0) = 8

\frac{1}{( s ^{2} + 2 ^{2} )}

\int x^{2} - 6 x d x

- \frac{1}{3} x^{2}

What is the d/(dθ)(5/(sin(θ)-2cos(θ))) ?
The d/(dθ)(5/(sin(θ)-2cos(θ))) is -(5(cos(θ)+2sin(θ)))/((sin(θ)-2cos(θ))^2)
What is the first d/(dθ)(5/(sin(θ)-2cos(θ))) ?
The first d/(dθ)(5/(sin(θ)-2cos(θ))) is -(5(cos(θ)+2sin(θ)))/((sin(θ)-2cos(θ))^2)