Find the derivative of the following functions from first principle: (i) -x (ii) (-x)⁻¹ (iii) sin(x + 1) (iv)

Miscellaneous Exercise · Q1

Q. Find the derivative of the following functions from first principle:

(i)

-x

(ii)

(-x)^{-1}

(iii)

\sin(x + 1)

(iv)

\cos\left(x - \dfrac{\pi}{8}\right)

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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:

Derivatives: $\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$
Physics: average velocity → instantaneous velocity
Economics: average cost change → marginal cost
Any field that studies how things change

Every time you see a derivative, you're looking at the limit of a difference quotient. Master this one expression, and you've unlocked the core idea of differential calculus.

Watch out

A common mistake: forgetting that $h$ is the change in the input, not the output. The numerator $f(x+h) - f(x)$ is the change in the output. Keep them straight: $\frac{\Delta \text{output}}{\Delta \text{input}}$ .

The difference quotient is the direct precursor to the formal definition of a derivative in the NCERT Class 11 Mathematics chapter on Limits and Derivatives, and "difference quotient formula and examples" is a commonly searched topic for CBSE board and JEE Main preparation. Understanding it as the slope of a secant line is essential groundwork for the "limits and derivatives important questions" that build up to differentiation in Class 12.

Concept: First principle of derivative: $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$ . For $f(x) = -x$ : $f'(x) = \lim_{h \to 0} \frac{-(x+h) - (-x)}{h} = \lim_{h \to 0} \frac{-h}{h} = -1.$
✓Final answer
$\boxed{-1}$
Concept: First principle with $f(x) = (-x)^{-1} = -\frac{1}{x}$ . $f'(x) = \lim_{h \to 0} \frac{-\frac{1}{x+h} + \frac{1}{x}}{h} = \lim_{h \to 0} \frac{-(x) + (x+h)}{h \cdot x(x+h)} = \lim_{h \to 0} \frac{h}{h \cdot x(x+h)} = \frac{1}{x^2}.$
✓Final answer
$\boxed{\frac{1}{x^2}}$
Concept: First principle for $\sin(x+1)$ using $\sin A - \sin B = 2\cos\frac{A+B}{2}\sin\frac{A-B}{2}$ . $f'(x) = \lim_{h \to 0} \frac{\sin(x+h+1) - \sin(x+1)}{h} = \lim_{h \to 0} \frac{2\cos\left(x+1+\frac{h}{2}\right) \sin\frac{h}{2}}{h} = \cos(x+1).$
✓Final answer
$\boxed{\cos(x+1)}$
Concept: First principle for $\cos\left(x - \frac{\pi}{8}\right)$ using $\cos A - \cos B = -2\sin\frac{A+B}{2}\sin\frac{A-B}{2}$ . $f'(x) = \lim_{h \to 0} \frac{\cos\left(x+h-\frac{\pi}{8}\right) - \cos\left(x-\frac{\pi}{8}\right)}{h} = \lim_{h \to 0} \frac{-2\sin\left(x-\frac{\pi}{8}+\frac{h}{2}\right) \sin\frac{h}{2}}{h} = -\sin\left(x-\frac{\pi}{8}\right).$
✓Final answer
$\boxed{-\sin\left(x-\frac{\pi}{8}\right)}$

The derivative from first principles uses the limit definition $f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$ . For (i) $-x$ , the derivative is $-1$ ; for (ii) $(-x)^{-1}$ , it is $\frac{1}{x^2}$ ; for (iii) $\sin(x+1)$ , it is $\cos(x+1)$ ; for (iv) $\cos\left(x - \frac{\pi}{8}\right)$ , it is $-\sin\left(x - \frac{\pi}{8}\right)$ .

The first principle of differentiation is the very definition of a derivative. It tells us that the derivative of a function $f(x)$ at a point $x$ is the limit of the slope of the secant line as the two points get infinitely close. Formally:

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

This is the foundation of all calculus. Every derivative rule — product rule, chain rule, quotient rule — is derived from this single limit. So when a problem asks you to find a derivative "from first principle," it means you must use this limit directly, without any shortcuts.

Let's work through each function one by one.

(i) $f(x) = -x$

Step 1: Write the definition.

We need:

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

Step 2: Substitute the function.

$f(x+h) = -(x+h) = -x - h$ and $f(x) = -x$ . So:

\frac{f(x+h) - f(x)}{h} = \frac{(-x - h) - (-x)}{h} = \frac{-x - h + x}{h} = \frac{-h}{h}

Step 3: Simplify and take the limit.

For $h \neq 0$ , $\frac{-h}{h} = -1$ . This is constant — it doesn't depend on $h$ at all. So:

\lim_{h \to 0} (-1) = -1

Watch out

A common mistake here is to forget that $h$ cancels completely before taking the limit. If you try to plug $h=0$ directly into $\frac{-h}{h}$ , you get $\frac{0}{0}$ , which is indeterminate. Always simplify first.

Step 4: State the result.

The derivative of $-x$ is $-1$ . This makes perfect sense: the graph of $y = -x$ is a straight line with slope $-1$ , so its derivative (the slope at every point) is constant $-1$ .

✓Final answer

-x

(ii) $f(x) = (-x)^{-1}$

First, note that $(-x)^{-1} = \frac{1}{-x} = -\frac{1}{x}$ , provided $x \neq 0$ . We'll use the form $f(x) = \frac{1}{-x}$ .

Step 1: Write the definition.

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

Step 2: Substitute.

$f(x+h) = \frac{1}{-(x+h)} = -\frac{1}{x+h}$ and $f(x) = -\frac{1}{x}$ . So:

\frac{f(x+h) - f(x)}{h} = \frac{-\frac{1}{x+h} + \frac{1}{x}}{h}

Step 3: Combine the numerator into a single fraction.

The numerator is $-\frac{1}{x+h} + \frac{1}{x} = \frac{-x + (x+h)}{x(x+h)} = \frac{h}{x(x+h)}$ .

So the whole expression becomes:

\frac{\frac{h}{x(x+h)}}{h} = \frac{h}{x(x+h)} \cdot \frac{1}{h} = \frac{1}{x(x+h)}

Step 4: Take the limit as $h \to 0$ .

\lim_{h \to 0} \frac{1}{x(x+h)} = \frac{1}{x(x+0)} = \frac{1}{x^2}

Tip

Notice that the derivative of $\frac{1}{-x}$ came out positive $\frac{1}{x^2}$ , not negative. This is because the negative sign in the denominator flips the usual derivative of $\frac{1}{x}$ (which is $-\frac{1}{x^2}$ ). Always check signs carefully.

✓Final answer

(-x)^{-1}

(iii) $f(x) = \sin(x + 1)$

Step 1: Write the definition.

f'(x) = \lim_{h \to 0} \frac{\sin(x + h + 1) - \sin(x + 1)}{h}

Step 2: Use the sine difference identity.

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

Here, $A = x + h + 1$ and $B = x + 1$ . So:

\sin(x+h+1) - \sin(x+1) = 2 \cos\left(\frac{(x+h+1)+(x+1)}{2}\right) \sin\left(\frac{(x+h+1)-(x+1)}{2}\right)

Simplify the arguments:

$\frac{A+B}{2} = \frac{2x + h + 2}{2} = x + \frac{h}{2} + 1$
$\frac{A-B}{2} = \frac{h}{2}$

So the difference becomes:

2 \cos\left(x + \frac{h}{2} + 1\right) \sin\left(\frac{h}{2}\right)

Step 3: Substitute back into the limit.

\frac{\sin(x+h+1) - \sin(x+1)}{h} = \frac{2 \cos\left(x + \frac{h}{2} + 1\right) \sin\left(\frac{h}{2}\right)}{h}

Step 4: Rewrite to use the standard limit $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$ .

Multiply numerator and denominator by $\frac{1}{2}$ :

= \cos\left(x + \frac{h}{2} + 1\right) \cdot \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}}

Step 5: Take the limit as $h \to 0$ .

As $h \to 0$ , $\frac{h}{2} \to 0$ , so $\frac{\sin(h/2)}{h/2} \to 1$ . Also, $\cos\left(x + \frac{h}{2} + 1\right) \to \cos(x + 1)$ .

Therefore:

f'(x) = \cos(x + 1)

Important

The derivative of $\sin(\text{anything})$ is $\cos(\text{anything})$ times the derivative of "anything" (chain rule). Here, the "anything" is $x+1$ , whose derivative is $1$ , so the result is just $\cos(x+1)$ . The first-principles derivation confirms this.

✓Final answer

\sin(x+1)

(iv) $f(x) = \cos\left(x - \frac{\pi}{8}\right)$

Step 1: Write the definition.

f'(x) = \lim_{h \to 0} \frac{\cos\left(x + h - \frac{\pi}{8}\right) - \cos\left(x - \frac{\pi}{8}\right)}{h}

Step 2: Use the cosine difference identity.

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

Here, $A = x + h - \frac{\pi}{8}$ and $B = x - \frac{\pi}{8}$ . So:

\cos A - \cos B = -2 \sin\left(\frac{(x+h-\frac{\pi}{8})+(x-\frac{\pi}{8})}{2}\right) \sin\left(\frac{(x+h-\frac{\pi}{8})-(x-\frac{\pi}{8})}{2}\right)

Simplify the arguments:

$\frac{A+B}{2} = \frac{2x + h - \frac{\pi}{4}}{2} = x + \frac{h}{2} - \frac{\pi}{8}$
$\frac{A-B}{2} = \frac{h}{2}$

So the difference becomes:

-2 \sin\left(x + \frac{h}{2} - \frac{\pi}{8}\right) \sin\left(\frac{h}{2}\right)

Step 3: Substitute back into the limit.

\frac{\cos(x+h-\frac{\pi}{8}) - \cos(x-\frac{\pi}{8})}{h} = \frac{-2 \sin\left(x + \frac{h}{2} - \frac{\pi}{8}\right) \sin\left(\frac{h}{2}\right)}{h}

Step 4: Rewrite to use $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$ .

= -\sin\left(x + \frac{h}{2} - \frac{\pi}{8}\right) \cdot \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}}

Step 5: Take the limit as $h \to 0$ .

As $h \to 0$ , $\frac{\sin(h/2)}{h/2} \to 1$ , and $\sin\left(x + \frac{h}{2} - \frac{\pi}{8}\right) \to \sin\left(x - \frac{\pi}{8}\right)$ .

Therefore:

f'(x) = -\sin\left(x - \frac{\pi}{8}\right)

Watch out

A common pitfall: forgetting the negative sign in the cosine difference identity. The identity $\cos A - \cos B = -2 \sin(\frac{A+B}{2}) \sin(\frac{A-B}{2})$ has a minus sign in front. If you use $\cos A - \cos B = 2 \sin(\frac{A+B}{2}) \sin(\frac{B-A}{2})$ instead, you'll get the same result, but the sign is easy to mess up. Always double-check.

✓Final answer

\cos\left(x - \frac{\pi}{8}\right)

←Previous question · misc-examplesExample 22 — Find the derivative of (i) (x⁵ - cos x)/(sin x) (ii) (x + cos x)/(tan …→Next question · misc-exerciseQ2 — Find the derivative of (x + a).

Q.Find the derivative of the following functions from first principle:

(i) f(x)=−xf(x) = -xf(x)=−x

(ii) f(x)=(−x)−1f(x) = (-x)^{-1}f(x)=(−x)−1

(iii) f(x)=sin⁡(x+1)f(x) = \sin(x + 1)f(x)=sin(x+1)

(iv) f(x)=cos⁡(x−π8)f(x) = \cos\left(x - \frac{\pi}{8}\right)f(x)=cos(x−8π​)

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(i) $f(x) = -x$

(ii) $f(x) = (-x)^{-1}$

(iii) $f(x) = \sin(x + 1)$

(iv) $f(x) = \cos\left(x - \frac{\pi}{8}\right)$

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