What is the derivative of f(θ)=(sec(θ))/(1+sec(θ)) ?

The derivative of f(θ)=(sec(θ))/(1+sec(θ)) is (sin(θ))/((1+cos(θ))^2)

What is the first derivative of f(θ)=(sec(θ))/(1+sec(θ)) ?

The first derivative of f(θ)=(sec(θ))/(1+sec(θ)) is (sin(θ))/((1+cos(θ))^2)

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derivative of

f (θ) = \frac{sec ( θ )}{1 + sec ( θ )}

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

Solution steps

\frac{d}{d θ} (\frac{sec ( θ )}{1 + sec ( θ )})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

\frac{d}{d θ} (sec (θ)) = sec (θ) tan (θ)

\frac{d}{d θ} (1 + sec (θ)) = sec (θ) tan (θ)

= \frac{sec ( θ ) tan ( θ ) ( 1 + sec ( θ ) ) - sec ( θ ) tan ( θ ) sec ( θ )}{( 1 + sec ( θ ) ) ^{2}}

Simplify

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

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

f (x) = 4 x^{2} + 7 x

\int \frac{1}{u ^{2} - 4} d u

\frac{d}{d θ} (θ)

\int sin (7 x) cos (7 x) d x

\frac{\partial}{\partial x} (\frac{x - y}{2 x y + 5})

What is the derivative of f(θ)=(sec(θ))/(1+sec(θ)) ?
The derivative of f(θ)=(sec(θ))/(1+sec(θ)) is (sin(θ))/((1+cos(θ))^2)
What is the first derivative of f(θ)=(sec(θ))/(1+sec(θ)) ?
The first derivative of f(θ)=(sec(θ))/(1+sec(θ)) is (sin(θ))/((1+cos(θ))^2)