Question

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

Answer 1

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

& \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{aligned}$$ Now substituting the above derivatives in quotient rule, we get $$\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 $$\begin{aligned} & \Rightarrow \dfrac{dy}{d\theta }=\dfrac{\left( \sec \theta \tan \theta \right)\left( \left( 1+\sec \theta \right)-\sec \theta \right)}{{{\left( 1+\sec \theta \right)}^{2}}} \\\ & \Rightarrow \dfrac{dy}{d\theta }=\dfrac{\left( \sec \theta \tan \theta \right)\left( 1 \right)}{{{\left( 1+\sec \theta \right)}^{2}}} \\\ & \Rightarrow \dfrac{dy}{d\theta }=\dfrac{\left( \sec \theta \tan \theta \right)}{{{\left( 1+\sec \theta \right)}^{2}}} \\\ \end{aligned}$$ 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.