What is the d/(dθ)(5/(2cos(θ)+7sin(θ))) ?

The d/(dθ)(5/(2cos(θ)+7sin(θ))) is -(5(-2sin(θ)+7cos(θ)))/((2cos(θ)+7sin(θ))^2)

What is the first d/(dθ)(5/(2cos(θ)+7sin(θ))) ?

The first d/(dθ)(5/(2cos(θ)+7sin(θ))) is -(5(-2sin(θ)+7cos(θ)))/((2cos(θ)+7sin(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{5}{2 cos ( θ ) + 7 sin ( θ )})

- \frac{5 ( - 2 sin ( θ ) + 7 cos ( θ ) )}{( 2 cos ( θ ) + 7 sin ( θ ) ) ^{2}}

Solution steps

\frac{d}{d θ} (\frac{5}{2 cos ( θ ) + 7 sin ( θ )})

Take the constant out:

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

= 5 \frac{d}{d θ} (\frac{1}{2 cos ( θ ) + 7 sin ( θ )})

Apply exponent rule:

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

= 5 \frac{d}{d θ} ((2 cos (θ) + 7 sin (θ))^{- 1})

Apply the chain rule:

- \frac{1}{( 2 cos ( θ ) + 7 sin ( θ ) ) ^{2}} \frac{d}{d θ} (2 cos (θ) + 7 sin (θ))

= - \frac{1}{( 2 cos ( θ ) + 7 sin ( θ ) ) ^{2}} \frac{d}{d θ} (2 cos (θ) + 7 sin (θ))

\frac{d}{d θ} (2 cos (θ) + 7 sin (θ)) = - 2 sin (θ) + 7 cos (θ)

= 5 (- \frac{1}{( 2 cos ( θ ) + 7 sin ( θ ) ) ^{2}} (- 2 sin (θ) + 7 cos (θ)))

Simplify

5 (- \frac{1}{( 2 cos ( θ ) + 7 sin ( θ ) ) ^{2}} (- 2 sin (θ) + 7 cos (θ))) : - \frac{5 ( - 2 sin ( θ ) + 7 cos ( θ ) )}{( 2 cos ( θ ) + 7 sin ( θ ) ) ^{2}}

= - \frac{5 ( - 2 sin ( θ ) + 7 cos ( θ ) )}{( 2 cos ( θ ) + 7 sin ( θ ) ) ^{2}}

\frac{d y}{d x} = \frac{( x + 1 )}{y}

\frac{\partial}{\partial y} (x y e^{- 5 y})

\frac{d}{d x} (\frac{16}{x ^{3}})

5 y^{^{'}} + y = - 3

f (x) = 3 x^{4} - 8 x^{3} + 2

What is the d/(dθ)(5/(2cos(θ)+7sin(θ))) ?
The d/(dθ)(5/(2cos(θ)+7sin(θ))) is -(5(-2sin(θ)+7cos(θ)))/((2cos(θ)+7sin(θ))^2)
What is the first d/(dθ)(5/(2cos(θ)+7sin(θ))) ?
The first d/(dθ)(5/(2cos(θ)+7sin(θ))) is -(5(-2sin(θ)+7cos(θ)))/((2cos(θ)+7sin(θ))^2)