Differentiation – definitions, rules, implicit differentiation, applications

1. What is a derivative?

The derivative of a function $f$ at a point $x=a$ is the limit

$$f^{\prime }(a)=\underset{h\to 0}{\lim }\frac{f(a+h)-f(a)}{h}$$

provided the limit exists. Geometrically it is the slope of the tangent line to the curve $y=f(x)$ at $(a,f(a))$.

Key idea: The derivative converts a static relationship into a dynamic one – it tells how fast the output changes when the input changes by an infinitesimal amount.

2. Basic differentiation rules

flowchart LR
    A["Basic rules"] --> B["Constant"]
    A --> C["Power"]
    A --> D["Constant multiple"]
    A --> E["Sum/Difference"]
    A --> F["Product"]
    A --> G["Quotient"]
    A --> H["Chain"]

Example 1 – Using the power and constant‑multiple rules

Find $\frac{d}{dx}(5x^4-3x^2+7)$.

$$\begin{array}{rl}\frac{d}{dx}(5x^4)& =5\cdot 4x^3=20x^3\\ \frac{d}{dx}(-3x^2)& =-3\cdot 2x=-6x\\ \frac{d}{dx}(7)& =0\end{array}$$

Hence $\frac{d}{dx}(5x^4-3x^2+7)=20x^3-6x$.

3. Implicit differentiation

When a relation between $x$ and $y$ is given implicitly (e.g. $x^2+y^2=25$), we differentiate both sides with respect to $x$, treating $y$ as a function of $x$ (i.e. $y=y(x)$). Apply the chain rule to every term containing $y$.

Worked example (exam style)

Problem: Define implicit differentiation and find the slope of the circle $x^2+y^2=25$ at the point $(3,-4)$.

Solution:

1. Differentiate both sides w.r.t. $x$:

$$\frac{d}{dx}(x^2)+\frac{d}{dx}(y^2)=\frac{d}{dx}(25)$$

$$2x+2y\frac{dy}{dx}=0\text{(since }\frac{d}{dx}(y^2)=2y\frac{dy}{dx}\text{)}$$

2. Solve for $\frac{dy}{dx}$:

$$\frac{dy}{dx}=-\frac{x}{y}$$

3. Substitute the point $(3,-4)$:

$${\frac{dy}{dx}|}_{(3,-4)}=-\frac{3}{-4}=\frac{3}{4}$$

Thus the slope of the tangent line at $(3,-4)$ is $\frac{3}{4}$.

The tangent line equation can be written as

$$y+4=\frac{3}{4}(x-3)$$

4. Differentiation of trigonometric, exponential and logarithmic functions

Example 2 – Chain rule with trig‑exponential

Differentiate $f(x)=e^{\sin x}$.

$$f^{\prime }(x)=e^{\sin x}\cdot \cos x$$

5. Higher‑order derivatives

The second derivative $f^{\prime \prime }(x)$ is the derivative of $f^{\prime }(x)$. It measures curvature or acceleration.

• If $f^{\prime \prime }(x)>0$ on an interval, the graph is concave up (shaped like a cup).
• If $f^{\prime \prime }(x)<0$, the graph is concave down.

Example 3 – Motion interpretation

A car’s position is $s(t)=4t^3-3t^2+2t$ (metres).

$$\begin{array}{rl}v(t)& =s^{\prime }(t)=12t^2-6t+2\text{(velocity)}\\ a(t)& =v^{\prime }(t)=s^{\prime \prime }(t)=24t-6\text{(acceleration)}\end{array}$$

At $t=1$ s, $v(1)=8$ m/s and $a(1)=18$ m/s${}^2$.

6. Applications of differentiation

Example 4 – Optimization (max profit)

A small shop sells $x$ units per day, earning profit $P(x)=-2x^2+120x-500$.

$$P^{\prime }(x)=-4x+120$$

Set $P^{\prime }(x)=0$ → $x=30$.

Second derivative $P^{\prime \prime }(x)=-4<0$ → maximum.

Maximum profit occurs when 30 units are sold per day.

7. Real‑world connections

In the real world

1. eSewa transaction fee – eSewa charges a fee that follows a piecewise linear cost function:

   $$C(A)=\{\begin{array}{ll}10\text{ NPR}& A\le 500\\ 0.02A+9& A>500\end{array}$$

   The marginal fee $\frac{dC}{dA}=0.02$ NPR per additional rupee for amounts above 500 NPR. This derivative tells the system how much extra charge to add for each extra rupee transferred, enabling real‑time fee calculation.

2. Daraz order‑processing queue – Daraz models the number of orders processed per hour as $O(t)=120-5t$ (orders/hour, where $t$ is hours after a peak). The rate of change $\frac{dO}{dt}=-5$ shows the queue is slowing down by 5 orders each hour, helping the logistics team allocate extra staff before the queue collapses.

3. Bank loan interest (compound) – For a loan amount $L(t)=L_0{e}^{rt}$ (continuous compounding, rate $r$), the derivative $L^{\prime }(t)=r{L}_0{e}^{rt}=rL(t)$ gives the instantaneous interest accrued at any moment. A bank uses this to compute the exact interest to credit to the borrower’s account each day.

Worked‑example tie‑in: The bank’s interest formula is identical to the exponential differentiation rule $\frac{d}{dx}{e}^{kx}=k{e}^{kx}$. Understanding this rule lets a student instantly write the marginal interest without re‑deriving it.

8. Summary of differentiation rules (quick reference)

classDiagram
    class "Basic Rules"{
        +Constant
        +Power
        +ConstantMultiple
        +SumDifference
    }
    class Product{
        +Rule: (fg)'=f'g+fg'
    }
    class Quotient{
        +Rule: (f/g)'=(f'g-fg')/g^2
    }
    class Chain{
        +Rule: (f∘g)'=f'(g)·g'
    }
    BasicRules <|-- Product
    BasicRules <|-- Quotient
    BasicRules <|-- Chain

9. Common pitfalls and how to avoid them

Exam tip

• Read the question carefully: if a point is given, substitute before simplifying the derivative; this avoids algebraic mistakes (as in the implicit‑circle problem).
• Write the rule name (e.g., “using the chain rule”) before the algebra; examiners award marks for method.
• For related‑rates problems, list all variables, write a relation, differentiate implicitly, then plug in the known rates.
• Check units: derivatives have units of “output per input” (e.g., m/s for position vs. time). A mismatched unit signals an error.
• Time‑saving shortcut: memorize the standard derivatives of $\sin ,\cos ,e^x,\ln x$; they appear in >70 % of questions.

Tangent line illustrating the derivative as slope (Image: Jacj at English Wikipedia / Later versions were uploaded by , Public domain, via Wikimedia Commons)