Question Statement

Find the derivative of the function:

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

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Background and Explanation

This problem combines the chain rule (for the outer squaring function) with the quotient rule (for the inner rational function). Recognizing that $f(x)$ is a composition of functions—where the outer function is $u^2$ and the inner function is $u=\frac{x^2-1}{x^2+1}$—allows us to differentiate layer by layer.

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Solution

We begin by identifying $f(x)$ as a composite function. Let $u=\frac{x^2-1}{x^2+1}$, so that $f(x)=u^2$.

Step 1: Apply the Chain Rule

Using the chain rule $\frac{d}{dx}[u^2]=2u\cdot \frac{du}{dx}$:

$f^{\prime }(x)=2{\left(\frac{x^2-1}{x^2+1}\right)}^{2-1}\cdot \frac{d}{dx}\left(\frac{x^2-1}{x^2+1}\right)$

$f^{\prime }(x)=2\left(\frac{x^2-1}{x^2+1}\right)\cdot \frac{d}{dx}\left(\frac{x^2-1}{x^2+1}\right)$

Step 2: Differentiate the Inner Function (Quotient Rule)

To find $\frac{d}{dx}\left(\frac{x^2-1}{x^2+1}\right)$, we apply the quotient rule $\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v{u}^{\prime }-u{v}^{\prime }}{v^2}$ where $u=x^2-1$ and $v=x^2+1$:

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

Computing the derivatives in the numerator:

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

Substituting these back:

$=\frac{(x^2+1)(2x)-(x^2-1)(2x)}{(x^2+1{)}^2}$

Step 3: Simplify the Numerator

Expanding the terms in the numerator:

• $(x^2+1)(2x)=2x^3+2x$
• $(x^2-1)(2x)=2x^3-2x$

Therefore: $(2x^3+2x)-(2x^3-2x)=2x^3+2x-2x^3+2x=4x$

So the derivative of the inner function is: $\frac{d}{dx}\left(\frac{x^2-1}{x^2+1}\right)=\frac{4x}{(x^2+1{)}^2}$

Step 4: Combine Results

Substituting back into our chain rule expression from Step 1:

$f^{\prime }(x)=2\left(\frac{x^2-1}{x^2+1}\right)\cdot \frac{4x}{(x^2+1{)}^2}$

Multiplying the factors: $f^{\prime }(x)=\frac{2(x^2-1)\cdot 4x}{(x^2+1)(x^2+1{)}^2}$

$f^{\prime }(x)=\frac{8x(x^2-1)}{(x^2+1{)}^3}$

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Key Formulas or Methods Used

• Chain Rule: $\frac{d}{dx}[f(g(x))]=f^{\prime }(g(x))\cdot g^{\prime }(x)$
• Quotient Rule: $\frac{d}{dx}\left[\frac{u(x)}{v(x)}\right]=\frac{v(x)u^{\prime }(x)-u(x)v^{\prime }(x)}{[v(x){]}^2}$
• Power Rule: $\frac{d}{dx}[x^n]=n{x}^{n-1}$

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Summary of Steps

1. Identify structure: Recognize $f(x)$ as a composite function (outer: squaring, inner: rational expression)
2. Apply chain rule: Differentiate the outer function, keeping the inner function unchanged, then multiply by the derivative of the inner function
3. Apply quotient rule: Differentiate the rational function $\frac{x^2-1}{x^2+1}$ using $\frac{v{u}^{\prime }-u{v}^{\prime }}{v^2}$
4. Simplify: Expand and combine like terms in the numerator to obtain $4x$
5. Combine factors: Multiply the constants and combine the denominator terms to get the final result $\frac{8x(x^2-1)}{(x^2+1{)}^3}$