Differentiate with respect to $x$: $\dfrac{x^4 + x^3 + x^2 + 1}{x}$.

Differentiate with respect to x: (x⁴ + x³ + x² + 1)/(x).

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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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The derivative of $\frac{x^4 + x^3 + x^2 + 1}{x}$ is found by first simplifying the expression term-by-term (dividing each term by $x$ ) and then differentiating using the power rule. The result is $3x^2 + 2x + 1 - \frac{1}{x^2}$ .

When you see a rational function like this, your first instinct might be to reach for the quotient rule. That would work, but it's unnecessarily messy. The key insight here is that the denominator is just a single term — $x$ — so we can split the fraction into a sum of simpler terms. This turns a complicated quotient into a straightforward polynomial (plus one reciprocal term), which we can differentiate term-by-term using the power rule.

Let's walk through it.

Simplify the expression first. Divide each term in the numerator by $x$ :

\frac{x^4 + x^3 + x^2 + 1}{x} = \frac{x^4}{x} + \frac{x^3}{x} + \frac{x^2}{x} + \frac{1}{x}

Simplify each fraction:

= x^3 + x^2 + x + \frac{1}{x}

Notice that $\frac{1}{x}$ is $x^{-1}$ — we'll need that form for differentiation.

Differentiate term by term. The derivative of a sum is the sum of the derivatives. So:

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

Apply the power rule. Recall: $\frac{d}{dx}(x^n) = n x^{n-1}$ $\frac{d}{d x} (x^{n}) = n x^{n - 1}$ .
- For $x^3$ : $3x^{2}$
- For $x^2$ : $2x^{1} = 2x$
- For $x$ : $1 \cdot x^{0} = 1$ …

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Q.Differentiate with respect to xxx: x4+x3+x2+1x\dfrac{x^4 + x^3 + x^2 + 1}{x}xx4+x3+x2+1​.

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Q.Differentiate with respect to $x$ : $\dfrac{x^4 + x^3 + x^2 + 1}{x}$ .

Unlock everything free for 14 days