Find the derivative of $\dfrac{x}{1 + an x}$.

Question

Find the derivative of $\dfrac{x}{1 + 	an x}$.

Accepted Answer

> [!TLDR]
> Apply the quotient rule to $\frac{x}{1 + 	an x}$, treating $x$ as the numerator and $1 + 	an x$ as the denominator. The derivative is $\boxed{\frac{1 + 	an x - x\sec^2 x}{(1 + 	an x)^2}}$.

When you have one function divided by another, the quotient rule is your tool. The key insight is that differentiation distributes across division in a specific pattern: the derivative of $\frac{u}{v}$ depends on both how the top changes (while the bottom stays fixed) and how the bottom changes (which affects the whole fraction).

> [!FORMULA]
> $$\frac{d}{dx}\left(\frac{u}{v}ight) = \frac{v \cdot u' - u \cdot v'}{v^2}$$

The mnemonic "lo d-hi minus hi d-lo over lo-lo" captures this: bottom times derivative of top, minus top times derivative of bottom, all over bottom squared.

Let me identify the pieces and work through this systematically.

**1. Identify $u$ and $v$**

For $\frac{x}{1 + 	an x}$, we have:
- $u = x$, so $u' = 1$
- $v = 1 + 	an x$, so $v' = \sec^2 x$ (since the derivative of $	an x$ is $\sec^2 x$)

**2. Apply the quotient rule formula**

Substituting into $\frac{v \cdot u' - u \cdot v'}{v^2}$:

$$\frac{d}{dx}\left(\frac{x}{1 + 	an x}ight) = \frac{(1 + 	an x)(1) - (x)(\sec^2 x)}{(1 + 	an x)^2}$$

**3. Simplify the numerator**

The numerator becomes:
$$(1 + 	an x) - x\sec^2 x = 1 + 	an x - x\sec^2 x$$ …

Q.Find the derivative of $\dfrac{x}{1 + \tan x}$ .

Concept understanding — Rules of Differentiation

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Q.Find the derivative of $\dfrac{x}{1 + \tan x}$ .

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