2.3 Q-8

Question Statement

Differentiate the following expression with respect to $x$:
$y = \frac{\left(x^2 + 1\right)^2}{x^2 - 1}$

---

Background and Explanation

This problem requires the application of the quotient rule for differentiation because the function is expressed as a ratio of two terms. Additionally, the chain rule is used to differentiate composite functions like $\left(x^2 + 1\right)^2$ and $x^2 - 1$.

Key Differentiation Rules:

1. Quotient Rule:
   If $y = \frac{u}{v}$, then:
   $\frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}$
2. Chain Rule:
   If $y = f(g(x))$, then:
   $\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$

---

Solution

Step 1: Identify the numerator and denominator

The given expression is:
$y = \frac{\left(x^2 + 1\right)^2}{x^2 - 1}$
Here:

• Numerator: $u = \left(x^2 + 1\right)^2$
• Denominator: $v = x^2 - 1$

Step 2: Apply the quotient rule

The quotient rule states:
$\frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}$

a) Differentiate $u = \left(x^2 + 1\right)^2$ using the chain rule:

• Outer function: $(\cdot)^2 \implies 2 \cdot (\cdot)$
• Inner function: $x^2 + 1 \implies \frac{d}{dx}(x^2 + 1) = 2x$

Thus: $\frac{du}{dx} = 2\left(x^2 + 1\right)(2x) = 4x\left(x^2 + 1\right)$

b) Differentiate $v = x^2 - 1$:

$\frac{dv}{dx} = 2x$

Step 3: Substitute into the quotient rule

Substituting the derivatives into the quotient rule formula:
$\frac{dy}{dx} = \frac{\left(x^2 - 1\right) \cdot 4x\left(x^2 + 1\right) - \left(x^2 + 1\right)^2 \cdot 2x}{\left(x^2 - 1\right)^2}$

Step 4: Simplify the numerator

Factor out $2x$: $\frac{dy}{dx} = \frac{2x \left[2\left(x^2 - 1\right)\left(x^2 + 1\right) - \left(x^2 + 1\right)^2\right]}{\left(x^2 - 1\right)^2}$

Expand each term:

1. $2\left(x^2 - 1\right)\left(x^2 + 1\right) = 2(x^4 - 1)$
2. $\left(x^2 + 1\right)^2 = x^4 + 2x^2 + 1$

So: $\text{Numerator} = 4x(x^4 - 1) - 2x(x^4 + 2x^2 + 1)$

Combine terms: $\text{Numerator} = 2x\left[(x^4 - 1) - (x^4 + 2x^2 + 1)\right]$
$= 2x\left(x^2 + 1\right)\left(x^2 + 3\right)$

Final Result:

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

---

Key Formulas or Methods Used

1. Quotient Rule:
   $\frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}$

2. Chain Rule:
   $\frac{d}{dx}\left[\left(f(x)\right)^n\right] = n \cdot f(x)^{n-1} \cdot f'(x)$

3. Simplifications using algebraic expansion and factoring.

---

Summary of Steps

1. Identify $u$ (numerator) and $v$ (denominator).
2. Differentiate $u = \left(x^2 + 1\right)^2$ using the chain rule: $\frac{du}{dx} = 4x\left(x^2 + 1\right)$
3. Differentiate $v = x^2 - 1$:
   $\frac{dv}{dx} = 2x$
4. Apply the quotient rule and simplify the numerator.
5. Factor the result for clarity: $\frac{dy}{dx} = \frac{2x\left(x^2 + 1\right)\left(x^2 + 3\right)}{\left(x^2 - 1\right)^2}$