how do you differentiate \[f\left( \theta \right)=\dfrac{\sec \theta }{1+\sec \theta }\]?

Hint: The quotient rule is a rule for differentiating expressions in which one function is divided by another function. The rule follows from the limit definition of derivative
and is given by \[\dfrac{dy}{dx}=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}\] if \[y=\dfrac{u}{v}\] where means derivative of y with respect to x, \[\dfrac{dv}{dx}\] means derivative of v with respect to x and \[\dfrac{du}{dx}\] means derivative of u with respect to x. The above formula is applicable only if u and v are differentiable.

Complete step by step answer:
As per the given question, we have to differentiate the given function using the product rule. Here, the given function to be differentiated is \[f\left( \theta \right)=\dfrac{\sec \theta }{1+\sec \theta }\].
Now, let \[y=f\left( \theta \right)\] then \[u=\sec \theta \] and \[v=1+\sec \theta \].
 We know that the derivative of constant is 0. That is \[\dfrac{d}{d\theta }\left( c \right)=0\]. The derivative of \[\sec \theta \] is \[\sec \theta \tan \theta \].
The derivative of function u: -
By comparing the function ‘u’ with the function \[\sec \theta \],function ‘u’ is the same as the\[\sec \theta \] function. Then,
\[\Rightarrow \]\[\dfrac{d}{d\theta }\left( \sec \theta \right)=\sec \theta \tan \theta \]
The derivative of function v: -
The function ‘v’ has both the functions \[\sec \theta \] and constant. Then derivative of v becomes
\[\begin{align}
  & \Rightarrow \dfrac{d}{d\theta }\left( 1+\sec \theta \right)=\dfrac{d}{d\theta }\left( 1 \right)+\dfrac{d}{d\theta }\left( \sec \theta \right) \\
 & \Rightarrow \dfrac{d}{d\theta }\left( 1+\sec \theta \right)=0+\sec \theta \tan \theta \\
 & \Rightarrow \dfrac{d}{d\theta }\left( 1+\sec \theta \right)=\sec \theta \tan \theta \\
\end{align}\]
Now substituting the above derivatives in quotient rule, we get
\[\Rightarrow \]\[\dfrac{dy}{d\theta }=\dfrac{v\dfrac{du}{d\theta }-u\dfrac{dv}{d\theta }}{{{v}^{2}}}\]
 \[\Rightarrow \]\[\dfrac{dy}{d\theta }=\dfrac{\left( 1+\sec \theta \right)\left( \sec \theta \tan \theta \right)-\sec \theta \left( \sec \theta \tan \theta \right)}{{{\left( 1+\sec \theta \right)}^{2}}}\]
 Since \[\sec \theta \tan \theta \] is common in both the terms. We take it common then
\[\begin{align}
  & \Rightarrow \dfrac{dy}{d\theta }=\dfrac{\left( \sec \theta \tan \theta \right)\left( \left( 1+\sec \theta \right)-\sec \theta \right)}{{{\left( 1+\sec \theta \right)}^{2}}} \\
 & \Rightarrow \dfrac{dy}{d\theta }=\dfrac{\left( \sec \theta \tan \theta \right)\left( 1 \right)}{{{\left( 1+\sec \theta \right)}^{2}}} \\
 & \Rightarrow \dfrac{dy}{d\theta }=\dfrac{\left( \sec \theta \tan \theta \right)}{{{\left( 1+\sec \theta \right)}^{2}}} \\
\end{align}\]
On solving the above equation, we get
\[\Rightarrow \]\[\dfrac{dy}{d\theta }=\dfrac{d\left( f\left( \theta \right) \right)}{d\theta }=\dfrac{\sec \theta \tan \theta }{{{\left( 1+\sec \theta \right)}^{2}}}\]
Therefore, the derivative of the given function \[f\left( \theta \right)=\dfrac{\sec \theta }{1+\sec \theta }\] is \[\dfrac{\sec \theta \tan \theta }{{{\left( 1+\sec \theta \right)}^{2}}}\].

Note: In order to solve these types of problems, we must have enough knowledge about derivatives of basic functions. The quotient rule is nothing but take the derivative of u multiplied by v and subtract u multiplied by the derivative of v and divide with \[{{v}^{2}}\]. We must avoid calculation mistakes to get the expected answers.