Differentiate with respect to x using first principle: xcos x.

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Concept understanding — Difference Quotient

The Difference Quotient: What It Is and Why It Matters

Imagine you're tracking the distance a car has travelled over time. At 2:00 PM, the odometer reads 40 km. At 2:30 PM, it reads 70 km. How fast was the car going on average during that half-hour?

You'd calculate: $\frac{70 - 40}{0.5 \text{ hours}} = 60$ km/h.

That fraction — change in distance divided by change in time — is the average rate of change. The difference quotient is just a formal, algebraic way of writing that same idea for any function.

The Intuition: Slope of a Secant Line

Take any function $f(x)$ . Pick two points on its graph: $(x, f(x))$ and $(x + h, f(x + h))$ , where $h$ is some horizontal step (positive or negative). The line that cuts through both points is called a secant line.

The slope of that secant line is:

\text{slope} = \frac{\text{rise}}{\text{run}} = \frac{f(x + h) - f(x)}{(x + h) - x} = \frac{f(x + h) - f(x)}{h}

That expression — $\frac{f(x + h) - f(x)}{h}$ — is the difference quotient.

Note

The name comes from "difference" (you subtract two function values) and "quotient" (you divide by $h$ ). It's literally a quotient of differences.

The Precise Statement

\frac{f(x + h) - f(x)}{h}, \quad h \neq 0

This gives the average rate of change of $f$ over the interval from $x$ to $x + h$ . Geometrically, it's the slope of the secant line through $(x, f(x))$ and $(x + h, f(x + h))$ .

Key restrictions:

$h$ cannot be zero (you can't divide by zero).
$x$ and $x + h$ must both be in the domain of $f$ .

A Concrete Example

Let $f(x) = x^2$ . Compute the difference quotient at $x = 3$ with $h = 0.1$ :

\frac{f(3 + 0.1) - f(3)}{0.1} = \frac{(3.1)^2 - 9}{0.1} = \frac{9.61 - 9}{0.1} = \frac{0.61}{0.1} = 6.1

This tells us: over the interval $[3, 3.1]$ , the function $x^2$ increases at an average rate of 6.1 units per unit change in $x$ .

If you shrink $h$ to 0.01, you'd get 6.01. As $h$ gets smaller, the average rate approaches 6 — which is exactly the instantaneous rate of change (the derivative) of $x^2$ at $x = 3$ .

Tip

The difference quotient is the bridge between average rates (which you can compute with simple algebra) and instantaneous rates (which require limits). When you take the limit as $h \to 0$ , you get the derivative.

Why You'll See It Everywhere

The difference quotient isn't just a classroom exercise. It's the foundation of calculus: …

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The first principle asks us to compute $\lim_{h \to 0} \frac{(x+h)\cos(x+h) - x\cos x}{h}$ by expanding the product and using standard limits. The derivative is $\boxed{\cos x - x\sin x}$ .

Why first principles?

The first principle of differentiation—also called the definition via limits—builds the derivative from the ground up. For any function $f(x)$ , we write

f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}.

This captures the instantaneous rate of change by shrinking the interval $h$ to zero. For $f(x) = x\cos x$ , a product of two functions, we could reach for the product rule—but the question demands we derive it from scratch using only the limit definition and standard trigonometric limits.

The key tools we'll need are:

$\lim_{h \to 0} \frac{\sin h}{h} = 1$
$\lim_{h \to 0} \frac{\cos h - 1}{h} = 0$

These are the bedrock limits of calculus, often proved geometrically or via squeeze theorem.

Step-by-step derivation

1. Set up the difference quotient.

Write $f(x) = x\cos x$ . Then

f'(x) = \lim_{h \to 0} \frac{(x+h)\cos(x+h) - x\cos x}{h}.

2. Expand the product in the numerator.

The term $(x+h)\cos(x+h)$ distributes to $x\cos(x+h) + h\cos(x+h)$ , so

f'(x) = \lim_{h \to 0} \frac{x\cos(x+h) + h\cos(x+h) - x\cos x}{h}.

Rearrange by grouping the $x$ terms:

= \lim_{h \to 0} \frac{x[\cos(x+h) - \cos x] + h\cos(x+h)}{h}.

3. Split the limit into two parts.

= \lim_{h \to 0} \left[ x \cdot \frac{\cos(x+h) - \cos x}{h} + \cos(x+h) \right].

The second term is straightforward: as $h \to 0$ , $\cos(x+h) \to \cos x$ . The first term requires more work.

4. Handle the cosine difference using the sum-to-product identity.

Recall the identity

\cos A - \cos B = -2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right).

Set $A = x+h$ and $B = x$ :

\cos(x+h) - \cos x = -2\sin\left(x + \frac{h}{2}\right)\sin\left(\frac{h}{2}\right).

\frac{\cos(x+h) - \cos x}{h} = -2\sin\left(x + \frac{h}{2}\right) \cdot \frac{\sin(h/2)}{h}.

5. Simplify the ratio $\frac{\sin(h/2)}{h}$ . …

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Q.Differentiate with respect to xxx using first principle: xcos⁡xx\cos xxcosx.

Why first principles?

Step-by-step derivation

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