Question

Differentiate the following function with respect to $x$: $\dfrac{\sec x+\tan x}{\sec x-\tan x}$

Answer 1

Simplify the expression by using the identity for the $\sec \theta$ and $\tan \theta$. After that we need to apply here the product rule of differentiation when the differentiating the two function which is multiplied together,
$\dfrac{d}{dx}\text{ }[f(x)\text{ }g(x)]\text{ }=\text{ }f(x)\text{ }\left[ \dfrac{d}{dx}\text{ }g(x) \right]\text{ }+\text{ }\left[ \dfrac{d}{dx}\text{ }f(x) \right]\text{ }g(x)$.
Then we need use the chain rule for differentiating function of function,
$\dfrac{d}{dx}f\left[ g(x) \right]=\dfrac{d}{dg(x)}f\left[ g(x) \right]\times \dfrac{d}{dx}g(x)$
By applying rules we have to apply the property of differentiation. Using these concepts we get the answer.

Complete step by step solution:
Let us first simplify the given expression.
Let's say that $y=\dfrac{\sec x+\tan x}{\sec x-\tan x}$.
Using $\sec \theta =\dfrac{1}{\cos \theta }$ and $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$, we get:
$\Rightarrow$ $y=\dfrac{\dfrac{1}{\cos x}+\dfrac{\sin x}{\cos x}}{\dfrac{1}{\cos x}-\dfrac{\sin x}{\cos x}}$
$\Rightarrow$ $y=\left( \dfrac{1+\sin x}{\cos x} \right)\times \left( \dfrac{\cos x}{1-\sin x} \right)$
$\Rightarrow$ $y=\dfrac{1+\sin x}{1-\sin x}$
Which can also be written as:
$\Rightarrow$ $y=(1+\sin x){{(1-\sin x)}^{-1}}$
Now, let us differentiate with respect to x.
Using the product rule of derivatives, we get:
$\Rightarrow$ $\dfrac{dy}{dx}=(1+\sin x)\left[ \dfrac{d}{dx}{{(1-\sin x)}^{-1}} \right]+\left[ \dfrac{d}{dx}(1+\sin x) \right]{{(1-\sin x)}^{-1}}$
Using the chain rule of derivatives, we get:
$\Rightarrow$ $\dfrac{dy}{dx}=(1+\sin x)\left[ (-1){{(1-\sin x)}^{-2}}\dfrac{d}{dx}(1-\sin x) \right]+\left[ \dfrac{d}{dx}(1+\sin x) \right]{{(1-\sin x)}^{-1}}$
Using $\dfrac{d}{dx}\sin x=\cos x$ and $\dfrac{d}{dx}k=0$, we get:
$\Rightarrow$ $\dfrac{dy}{dx}=(1+\sin x)(-1){{(1-\sin x)}^{-2}}(-\cos x)+\cos x{{(1-\sin x)}^{-1}}$
Which can be written as:
$\Rightarrow$ $\dfrac{dy}{dx}=\dfrac{(1+\sin x)(\cos x)}{{{(1-\sin x)}^{2}}}+\dfrac{\cos x}{(1-\sin x)}$
Separating the common factor $\cos x$, and adding by equating the denominators, we get:
$\Rightarrow$ $\dfrac{dy}{dx}=\cos x\left[ \dfrac{1+\sin x}{{{(1-\sin x)}^{2}}}+\dfrac{1-\sin x}{{{(1-\sin x)}^{2}}} \right]$

$\Rightarrow$ $\dfrac{dy}{dx}=\dfrac{2\cos x}{{{(1-\sin x)}^{2}}}$, which is the required answer.

Note: The final answer can be modified in terms of other trigonometric functions, the resulting value being the same.
Derivatives of Trigonometric Functions:
$\dfrac{d}{dx}\sin x=\cos x$ $\dfrac{d}{dx}\cos x=-\sin x$
$\dfrac{d}{dx}\tan x={{\sec }^{2}}x$ $\dfrac{d}{dx}\cot x=-{{\csc }^{2}}x$
$\dfrac{d}{dx}\sec x=\tan x\sec x$ $\dfrac{d}{dx}\csc x=-\cot x\csc x$