Differentiate with respect to $x$: $\left(x + \dfrac{1}{x}\right)^3$.

> [!TLDR] > To differentiate $\left(x + \dfrac{1}{x}\right)^3$, we apply the Chain Rule, treating $x + \dfrac{1}{x}$ as the inner function. The derivative is $3\left(x + \dfrac{1}{x}\right)^2 \left(1 - \dfrac{1}{x^2}\right)$. This problem asks us to differentiate a function that is composed of two simpler functions. We have an "outer" function, which is something cubed, and an "inner" function, which is $x + \frac{1}{x}$. When dealing with such composite functions, the **Chain Rule** is the essential tool. The Chain Rule allows us to find the derivative of $f(g(x))$ by first differentiating the outer function $f$ with respect to its argument $g(x)$, and then multiplying that result by the derivative of the inner function $g(x)$ with respect to $x$. It's like peeling an onion, layer by layer, and multiplying the derivatives of each layer. Let's apply this step-by-step. 1. **Identify the outer and inner functions.** Let the given function be $y = \left(x + \dfrac{1}{x}\right)^3$. We can define an inner function $u = x + \dfrac{1}{x}$. Then, the outer function becomes $y = u^3$. 2. **State the Chain Rule.** The Chain Rule states that if $y$ is a function of $u$, and $u$ is a function of $x$, then the derivative of $y$ with respect to $x$ is given by: > [!FORMULA] > $\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}$ 3. **Differentiate the outer function with respect to $u$.** We have $y = u^3$. Using the power rule for differentiation ($\frac{d}{dx}(x^n) = nx^{n-1}$): $\dfrac{dy}{du} = \dfrac{d}{du}(u^3) = 3u^{3-1} = 3u^2$. 4. **Differentiate the inner function with respect to $x$.** We have $u = x + \dfrac{1}{x}$. It's helpful to rewrite $\dfrac{1}{x}$ as $x^{-1}$. So, $u = x + x^{-1}$. Now, differentiate term by term: $\dfrac{du}{dx} = \dfrac{d}{dx}(x) + \dfrac{d}{dx}(x^{-1})$ The derivative of $x$ with respect to $x$ is $1$. The derivative of $x^{-1}$ with respect to $x$ (using the power rule) is

Differentiate with respect to x: (x + (1)/(x))³.

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Concept understanding — Rules of Differentiation

Computing every derivative directly from the limit definition ("from first principle") is correct but impractical for anything beyond the simplest functions. The five rules below, proved once from the definition, make differentiation a mechanical, algebraic process for any combination of known derivatives — no further limit computation is needed once they are established. Throughout, $u=u(x)$ , $v=v(x)$ are differentiable functions of $x$ , and $k$ is a constant.

Sum Rule (Theorem 10.2). $\dfrac{d}{dx}(u+v)=\dfrac{du}{dx}+\dfrac{dv}{dx}$ — differentiate term by term. Extends to any finite sum.

Product Rule (Theorem 10.3). $\dfrac{d}{dx}(uv) = u\dfrac{dv}{dx}+v\dfrac{du}{dx}$ , i.e. $(uv)'=uv'+vu'$ — "first times derivative of second, plus second times derivative of first." Extends to three or more factors, e.g. $(uvw)'=u'vw+uv'w+uvw'$ : differentiate one factor at a time, summing over which factor is differentiated.

Quotient Rule (Theorem 10.4). For $v\ne0$ , $\dfrac{d}{dx}\!\left(\dfrac{u}{v}\right) = \dfrac{v\,u'-u\,v'}{v^2}$ — "bottom times derivative of top, minus top times derivative of bottom, all over bottom squared."

Chain Rule (Theorem 10.5). If $y=f(u)$ and $u=g(x)$ are both differentiable, so $y=f(g(x))$ , then

\frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx} = f'(g(x))\,g'(x)

— differentiate the outer function $f$ with respect to its argument $u=g(x)$ (the inner function), then multiply by the derivative of the inner function. This is the single most-used rule in the chapter: every composite function (a power of an expression, $\sin$ of an expression, $e^{\text{expression}}$ , and so on) needs it.

Constant Multiple Rule (Theorem 10.6). $\dfrac{d}{dx}\big[k\,f(x)\big] = k\,f'(x)$ — a constant factor simply carries through differentiation unchanged. …

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To differentiate $\left(x + \dfrac{1}{x}\right)^3$ , we apply the Chain Rule, treating $x + \dfrac{1}{x}$ as the inner function. The derivative is $3\left(x + \dfrac{1}{x}\right)^2 \left(1 - \dfrac{1}{x^2}\right)$ .

This problem asks us to differentiate a function that is composed of two simpler functions. We have an "outer" function, which is something cubed, and an "inner" function, which is $x + \frac{1}{x}$ . When dealing with such composite functions, the Chain Rule is the essential tool.

The Chain Rule allows us to find the derivative of $f(g(x))$ by first differentiating the outer function $f$ with respect to its argument $g(x)$ , and then multiplying that result by the derivative of the inner function $g(x)$ with respect to $x$ . It's like peeling an onion, layer by layer, and multiplying the derivatives of each layer.

Let's apply this step-by-step.

Identify the outer and inner functions.

Let the given function be $y = \left(x + \dfrac{1}{x}\right)^3$ .

We can define an inner function $u = x + \dfrac{1}{x}$ .

Then, the outer function becomes $y = u^3$ .
State the Chain Rule.

The Chain Rule states that if $y$ is a function of $u$ , and $u$ is a function of $x$ , then the derivative of $y$ with respect to $x$ is given by:
$\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}$
Differentiate the outer function with respect to $u$ .

We have $y = u^3$ . Using the power rule for differentiation ( $\frac{d}{dx}(x^n) = nx^{n-1}$ ):

$\dfrac{dy}{du} = \dfrac{d}{du}(u^3) = 3u^{3-1} = 3u^2$ .
Differentiate the inner function with respect to $x$ .

We have $u = x + \dfrac{1}{x}$ .

It's helpful to rewrite $\dfrac{1}{x}$ as $x^{-1}$ . So, $u = x + x^{-1}$ .

Now, differentiate term by term:

$\dfrac{du}{dx} = \dfrac{d}{dx}(x) + \dfrac{d}{dx}(x^{-1})$

The derivative of $x$ with respect to $x$ is $1$ .

The derivative of $x^{-1}$ with respect to $x$ (using the power rule) is $-1 \cdot x^{-1-1} = -x^{-2} = -\dfrac{1}{x^2}$ .

So, $\dfrac{du}{dx} = 1 - \dfrac{1}{x^2}$ .
Combine the derivatives using the Chain Rule formula.

Substitute the expressions for $\dfrac{dy}{du}$ and $\dfrac{du}{dx}$ back into the Chain Rule: …

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Q.Differentiate with respect to $x$ : $\left(x + \dfrac{1}{x}\right)^3$ .

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