What is the d/(dθ)(2/(1-sin(θ))) ?

The d/(dθ)(2/(1-sin(θ))) is (2cos(θ))/((1-sin(θ))^2)

What is the first d/(dθ)(2/(1-sin(θ))) ?

The first d/(dθ)(2/(1-sin(θ))) is (2cos(θ))/((1-sin(θ))^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

x \to 5 lim (x^{2} + 5 x + 8)

\frac{x}{1 + 4 x ^{2}}

\frac{d}{d x} (e^{x} sin (x))

y = sin (9 x) + sin^{2} (9 x), (0, 0)

\frac{d y}{d x} + x^{2} sec (y) = 0, y (0) = 1

What is the d/(dθ)(2/(1-sin(θ))) ?
The d/(dθ)(2/(1-sin(θ))) is (2cos(θ))/((1-sin(θ))^2)
What is the first d/(dθ)(2/(1-sin(θ))) ?
The first d/(dθ)(2/(1-sin(θ))) is (2cos(θ))/((1-sin(θ))^2)