What is the d/(dθ)(1/(2cos(θ)+3sin(θ))) ?

The d/(dθ)(1/(2cos(θ)+3sin(θ))) is -(-2sin(θ)+3cos(θ))/((2cos(θ)+3sin(θ))^2)

What is the first d/(dθ)(1/(2cos(θ)+3sin(θ))) ?

The first d/(dθ)(1/(2cos(θ)+3sin(θ))) is -(-2sin(θ)+3cos(θ))/((2cos(θ)+3sin(θ))^2)

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\frac{d}{d θ} (\frac{1}{2 cos ( θ ) + 3 sin ( θ )})

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

Solution steps

\frac{d}{d θ} (\frac{1}{2 cos ( θ ) + 3 sin ( θ )})

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d}{d x} (sin (sin (2 x)))

\int - 2 d x

\int (3 x^{5} - 5 x + 6) d x

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

x \to 1 lim (\frac{∣ x - 1 ∣}{x - 1})

What is the d/(dθ)(1/(2cos(θ)+3sin(θ))) ?
The d/(dθ)(1/(2cos(θ)+3sin(θ))) is -(-2sin(θ)+3cos(θ))/((2cos(θ)+3sin(θ))^2)
What is the first d/(dθ)(1/(2cos(θ)+3sin(θ))) ?
The first d/(dθ)(1/(2cos(θ)+3sin(θ))) is -(-2sin(θ)+3cos(θ))/((2cos(θ)+3sin(θ))^2)