What is the d/(dθ)(6/(2-3sin(θ))) ?

The d/(dθ)(6/(2-3sin(θ))) is (18cos(θ))/((2-3sin(θ))^2)

What is the first d/(dθ)(6/(2-3sin(θ))) ?

The first d/(dθ)(6/(2-3sin(θ))) is (18cos(θ))/((2-3sin(θ))^2)

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\frac{d}{d θ} (\frac{6}{2 - 3 sin ( θ )})

\frac{18 cos ( θ )}{( 2 - 3 sin ( θ ) ) ^{2}}

Solution steps

\frac{d}{d θ} (\frac{6}{2 - 3 sin ( θ )})

Take the constant out:

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

= 6 \frac{d}{d θ} (\frac{1}{2 - 3 sin ( θ )})

Apply exponent rule:

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

= 6 \frac{d}{d θ} ((2 - 3 sin (θ))^{- 1})

Apply the chain rule:

- \frac{1}{( 2 - 3 sin ( θ ) ) ^{2}} \frac{d}{d θ} (2 - 3 sin (θ))

= - \frac{1}{( 2 - 3 sin ( θ ) ) ^{2}} \frac{d}{d θ} (2 - 3 sin (θ))

\frac{d}{d θ} (2 - 3 sin (θ)) = - 3 cos (θ)

= 6 (- \frac{1}{( 2 - 3 sin ( θ ) ) ^{2}} (- 3 cos (θ)))

Simplify

6 (- \frac{1}{( 2 - 3 sin ( θ ) ) ^{2}} (- 3 cos (θ))) : \frac{18 cos ( θ )}{( 2 - 3 sin ( θ ) ) ^{2}}

= \frac{18 cos ( θ )}{( 2 - 3 sin ( θ ) ) ^{2}}

2 y^{^{''}} + y = 0

\int (\frac{ln ^{2} ( x )}{x ^{\frac{5}{3}}}) d x

e^{2} x

\int sin^{3} (\frac{x}{6}) d x

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

What is the d/(dθ)(6/(2-3sin(θ))) ?
The d/(dθ)(6/(2-3sin(θ))) is (18cos(θ))/((2-3sin(θ))^2)
What is the first d/(dθ)(6/(2-3sin(θ))) ?
The first d/(dθ)(6/(2-3sin(θ))) is (18cos(θ))/((2-3sin(θ))^2)