Differentiate with respect to x using first principle: cos(x² + 1).

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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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First-principle differentiation means building the limit $\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$ from scratch. For $f(x) = \cos(x^2 + 1)$ , we use the cosine difference identity to simplify the numerator, then carefully take the limit to obtain $\boxed{-2x \sin(x^2 + 1)}$ .

Why first principles?

The definition of the derivative is the instantaneous rate of change, captured by the limit of the difference quotient. When we differentiate "from first principles," we're not allowed to use shortcut rules like the chain rule—we must return to the foundational limit and manipulate it algebraically until the limit can be evaluated.

For $f(x) = \cos(x^2 + 1)$ , the challenge is that cosine doesn't simplify nicely under addition in the argument. The key tool is the cosine difference identity:

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

This identity converts a difference of cosines into a product of sines, which behaves much better in limits.

Step-by-step derivation

1. Write the definition of the derivative.

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

For our function $f(x) = \cos(x^2 + 1)$ , substitute:

f'(x) = \lim_{h \to 0} \frac{\cos\left[(x+h)^2 + 1\right] - \cos(x^2 + 1)}{h}

2. Expand $(x+h)^2$ .

(x+h)^2 = x^2 + 2xh + h^2

So the numerator becomes:

\cos(x^2 + 2xh + h^2 + 1) - \cos(x^2 + 1)

3. Apply the cosine difference identity.

Let $A = x^2 + 2xh + h^2 + 1$ and $B = x^2 + 1$ . Then:

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

Compute the averages and differences:

\frac{A + B}{2} = \frac{(x^2 + 2xh + h^2 + 1) + (x^2 + 1)}{2} = \frac{2x^2 + 2xh + h^2 + 2}{2} = x^2 + xh + \frac{h^2}{2} + 1

\frac{A - B}{2} = \frac{(x^2 + 2xh + h^2 + 1) - (x^2 + 1)}{2} = \frac{2xh + h^2}{2} = xh + \frac{h^2}{2}

4. Substitute back into the difference quotient.

f'(x) = \lim_{h \to 0} \frac{-2 \sin\left(x^2 + xh + \frac{h^2}{2} + 1\right) \sin\left(xh + \frac{h^2}{2}\right)}{h}

Factor out the $-2$ :

f'(x) = -2 \lim_{h \to 0} \frac{\sin\left(x^2 + xh + \frac{h^2}{2} + 1\right) \sin\left(xh + \frac{h^2}{2}\right)}{h}

5. Separate the limit using the product rule for limits.

$f'(x) = -2 \left[\lim_{h \to 0} \sin\left(x^2 + xh + \frac{h^2}{2} + 1\right)\right] \left[\lim_{h \to 0} \frac{\sin\left(xh + \frac{h^2}{2}\right)}{h}\right]$ …

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Q.Differentiate with respect to xxx using first principle: cos⁡(x2+1)\cos(x^2 + 1)cos(x2+1).

Why first principles?

Step-by-step derivation

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Q.Differentiate with respect to $x$ using first principle: $\cos(x^2 + 1)$ .

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