Question Statement

Find $\frac{dy}{dx}$ given:

$y=\frac{\sec x}{1+\tan x}$

---

Background and Explanation

This problem requires applying the quotient rule for differentiation, along with the standard derivatives of trigonometric functions $\sec x$ and $\tan x$. You will also need to use the Pythagorean identity $1+{\tan }^{2}x={\sec }^{2}x$ (or equivalently ${\tan }^{2}x={\sec }^{2}x-1$) to simplify the final expression.

---

Solution

We recognize this function as a quotient of two functions. Let $u=\sec x$ and $v=1+\tan x$. We will apply the quotient rule:

$\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$

First, compute the derivatives of the numerator and denominator:

• $\frac{d}{dx}(\sec x)=\sec x\tan x$
• $\frac{d}{dx}(1+\tan x)=0+{\sec }^{2}x={\sec }^{2}x$

Now apply the quotient rule formula:

$\frac{dy}{dx}=\frac{(1+\tan x)\cdot (\sec x\tan x)-\sec x\cdot ({\sec }^{2}x)}{(1+\tan x{)}^2}$

Expand the numerator by distributing $(1+\tan x)(\sec x\tan x)$:

$\frac{dy}{dx}=\frac{\sec x\tan x+\sec x{\tan }^{2}x-{\sec }^{3}x}{(1+\tan x{)}^2}$

To simplify, use the identity ${\tan }^{2}x={\sec }^{2}x-1$ to replace ${\tan }^{2}x$:

$\frac{dy}{dx}=\frac{\sec x\tan x+\sec x({\sec }^{2}x-1)-{\sec }^{3}x}{(1+\tan x{)}^2}$

Distribute $\sec x$ in the middle term:

$\frac{dy}{dx}=\frac{\sec x\tan x+{\sec }^{3}x-\sec x-{\sec }^{3}x}{(1+\tan x{)}^2}$

Notice that the ${\sec }^{3}x$ and $-{\sec }^{3}x$ terms cancel each other:

$\frac{dy}{dx}=\frac{\sec x\tan x-\sec x}{(1+\tan x{)}^2}$

Finally, factor out $\sec x$ from the numerator to obtain the simplified result:

$\frac{dy}{dx}=\frac{\sec x(\tan x-1)}{(1+\tan x{)}^2}$

---

Key Formulas or Methods Used

• Quotient Rule: $\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\cdot u^{\prime }-u\cdot v^{\prime }}{v^2}$
• Derivative of secant: $\frac{d}{dx}(\sec x)=\sec x\tan x$
• Derivative of tangent: $\frac{d}{dx}(\tan x)={\sec }^{2}x$
• Pythagorean identity: ${\tan }^{2}x={\sec }^{2}x-1$ (derived from $1+{\tan }^{2}x={\sec }^{2}x$)

---

Summary of Steps

1. Identify components: Recognize $y$ as a quotient with $u=\sec x$ and $v=1+\tan x$
2. Differentiate components: Find $u^{\prime }=\sec x\tan x$ and $v^{\prime }={\sec }^{2}x$
3. Apply quotient rule: Substitute into $\frac{v{u}^{\prime }-u{v}^{\prime }}{v^2}$
4. Expand: Multiply out the terms in the numerator
5. Apply identity: Replace ${\tan }^{2}x$ with ${\sec }^{2}x-1$
6. Simplify: Expand and cancel the ${\sec }^{3}x$ terms
7. Factor: Factor out $\sec x$ to obtain the final simplified derivative $\frac{\sec x(\tan x-1)}{(1+\tan x{)}^2}$