Part 5 · Chapter 23

Differentiability & Differentiation

The slope of a curve at a single point — turning the limit of a secant into the derivative, and a handful of rules into a complete differentiation toolkit

Fundamentals of Mathematics Prof. Mithun Mondal Reading time ≈ 42 min
i What you'll learn
  • The derivative as a limit of secant slopes, and as the slope of the tangent.
  • The link between differentiability and continuity, via left- and right-hand derivatives.
  • The table of standard derivatives for powers, trigonometric, exponential and logarithmic functions.
  • The product, quotient and chain rules that combine them.
  • Implicit, logarithmic, parametric and inverse-function differentiation.
  • The inverse-trigonometric derivatives and higher-order derivatives.
Section 23-1

The Derivative as a Limit

Take two points on a curve and the line through them — a secant — has a definite slope. Slide the second point toward the first and the secant rotates into the tangent. The slope it settles on is the derivative, the instantaneous rate of change of \(f\) at that point.

P Q h secant tangent
As \(h\to0\), the secant slope tends to the tangent slope \(f'(x)\)
📉
The derivative from first principles
\(f'(x)=\dfrac{dy}{dx}=\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}\)

The quotient inside is the secant slope between \(x\) and \(x+h\); its limit, when it exists, is the slope of the tangent. The same quantity is the instantaneous rate of change of \(y\) with respect to \(x\).

Section 23-2

Differentiability & Continuity

A function is differentiable at \(a\) only if the one-sided derivatives agree. Every differentiable function is continuous — but the converse fails, as a sharp corner shows.

slope −1 slope +1 corner at 0
\(|x|\) is continuous at \(0\) but the one-sided slopes \(-1,+1\) disagree
🔗
Differentiability and its consequence
\(\text{LHD}=\text{RHD}\ \Rightarrow\ \text{differentiable};\qquad \text{differentiable}\ \Rightarrow\ \text{continuous}\)

The left- and right-hand derivatives are \(\lim_{h\to0^-}\tfrac{f(a+h)-f(a)}{h}\) and \(\lim_{h\to0^+}\tfrac{f(a+h)-f(a)}{h}\); the function is differentiable at \(a\) only if they are equal. Differentiability forces continuity, but continuity alone is not enough — \(|x|\) is continuous yet not differentiable at \(0\).

Section 23-3

Derivatives of Standard Functions

Differentiating from first principles once for each family yields a table that, combined with the rules ahead, covers almost everything you will meet.

🧰
The standard table
\(\dfrac{d}{dx}x^n=nx^{n-1},\quad \dfrac{d}{dx}e^x=e^x,\quad \dfrac{d}{dx}\ln x=\dfrac1x\)

Trigonometric: \((\sin x)'=\cos x\), \((\cos x)'=-\sin x\), \((\tan x)'=\sec^2x\). Exponential base \(a\): \((a^x)'=a^x\ln a\). And \((\sqrt{x})'=\tfrac{1}{2\sqrt{x}}\) is just the power rule with \(n=\tfrac12\).

Constant multiple

\(\dfrac{d}{dx}\big(c\,f\big)=c\,f'\) — constants pass straight through.

Sum & difference

\(\dfrac{d}{dx}\big(f\pm g\big)=f'\pm g'\) — differentiate term by term.

Constants vanish

\(\dfrac{d}{dx}(c)=0\) — a constant has zero rate of change.

Section 23-4

Product & Quotient Rules

The derivative of a product is not the product of the derivatives. Two dedicated rules handle products and quotients of functions.

✖️
Product and quotient rules
\((fg)'=f'g+fg';\qquad \left(\dfrac{f}{g}\right)'=\dfrac{f'g-fg'}{g^2}\)

Read the product rule as "derivative of the first times the second, plus the first times derivative of the second." For the quotient, the order in the numerator matters — it is \(f'g-fg'\), not the reverse — and the denominator is squared.

Section 23-5

The Chain Rule

For a composite function — a function of a function — differentiate the outer function at the inner one, then multiply by the derivative of the inner. This is the single most-used rule in all of calculus.

⛓️
The chain rule
\(\dfrac{d}{dx}f\big(g(x)\big)=f'\big(g(x)\big)\cdot g'(x)\)

In Leibniz notation, if \(y=f(u)\) and \(u=g(x)\), then \(\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot\dfrac{du}{dx}\) — the rates multiply. So \(\dfrac{d}{dx}\sin(3x^2+1)=\cos(3x^2+1)\cdot6x\).

Section 23-6

Implicit & Logarithmic Differentiation

When \(y\) is tangled into an equation with \(x\) rather than solved for, differentiate implicitly, treating \(y\) as a function of \(x\) and applying the chain rule to every \(y\) term. When the function is a tower of powers or a long product, take logarithms first.

🌀
Two specialised techniques
\(\text{implicit: differentiate both sides};\qquad \text{logarithmic: }\ln y\text{ first}\)

For an implicit relation, each \(y\) contributes a factor \(\tfrac{dy}{dx}\); collect those terms and solve. For logarithmic differentiation, take \(\ln\) of both sides to turn powers into products and products into sums, then differentiate — indispensable for \(y=x^x\) and similar.

Section 23-7

Parametric & Inverse Differentiation

When a curve is given by \(x=x(t),\ y=y(t)\), the slope comes from dividing the two rates. And the derivative of an inverse function is the reciprocal of the original derivative.

🔁
Parametric and inverse derivatives
\(\dfrac{dy}{dx}=\dfrac{\,dy/dt\,}{\,dx/dt\,};\qquad \dfrac{dy}{dx}=\dfrac{1}{\,dx/dy\,}\)

The parametric formula follows from the chain rule with parameter \(t\). The inverse-function rule says the slope of \(f^{-1}\) at a point is the reciprocal of the slope of \(f\) at the corresponding point — the graphs being reflections in \(y=x\).

! Inverse-trigonometric derivatives

These follow from the inverse rule: \(\dfrac{d}{dx}\sin^{-1}x=\dfrac{1}{\sqrt{1-x^2}}\), \(\dfrac{d}{dx}\cos^{-1}x=-\dfrac{1}{\sqrt{1-x^2}}\), \(\dfrac{d}{dx}\tan^{-1}x=\dfrac{1}{1+x^2}\), and \(\dfrac{d}{dx}\cot^{-1}x=-\dfrac{1}{1+x^2}\).

Section 23-8

Higher-Order Derivatives

Differentiating the derivative gives the second derivative, and so on. The second derivative measures how the slope itself is changing — the curvature that the next chapter exploits.

📐
Repeated differentiation
\(f''(x)=\dfrac{d}{dx}\big(f'(x)\big)=\dfrac{d^2y}{dx^2}\)

Each further differentiation drops the polynomial degree by one, so an \(n\)-th degree polynomial has zero \((n{+}1)\)-th derivative. The notations \(f'',\ y'',\ \tfrac{d^2y}{dx^2}\) all mean the same thing.

Worked Examples

Putting It to Work

1 From first principles

Problem. Differentiate \(f(x)=x^2\) using the definition.

Solution. Form the difference quotient and take the limit:

Working
\[ f'(x)=\lim_{h\to0}\frac{(x+h)^2-x^2}{h}=\lim_{h\to0}\frac{2xh+h^2}{h}=\lim_{h\to0}(2x+h)=2x \]
2 A function that is not differentiable

Problem. Show that \(f(x)=|x|\) is continuous but not differentiable at \(0\).

Solution. It is continuous since \(\lim_{x\to0}|x|=0=f(0)\). But the one-sided derivatives differ:

Working
\[ \text{LHD}=\lim_{h\to0^-}\frac{|h|}{h}=-1,\quad \text{RHD}=\lim_{h\to0^+}\frac{|h|}{h}=+1\ \Rightarrow\ \text{not differentiable} \]
3 The quotient rule

Problem. Differentiate \(y=\dfrac{x^2+1}{x-1}\).

Solution. With \(f=x^2+1,\ g=x-1\), apply \(\dfrac{f'g-fg'}{g^2}\):

Working
\[ y'=\frac{2x(x-1)-(x^2+1)(1)}{(x-1)^2}=\frac{x^2-2x-1}{(x-1)^2} \]
4 The chain rule

Problem. Differentiate \(y=\sin(3x^2+1)\).

Solution. Outer derivative \(\cos\) at the inner, times the inner's derivative \(6x\):

Working
\[ y'=\cos(3x^2+1)\cdot 6x=6x\cos(3x^2+1) \]
5 Logarithmic differentiation

Problem. Differentiate \(y=x^x\).

Solution. Take \(\ln\) of both sides, then differentiate implicitly:

Working
\[ \ln y=x\ln x\ \Rightarrow\ \frac{y'}{y}=\ln x+1\ \Rightarrow\ y'=x^x(\ln x+1) \]
6 Implicit differentiation

Problem. Find \(\dfrac{dy}{dx}\) for \(x^2+y^2=25\).

Solution. Differentiate both sides with respect to \(x\), using the chain rule on \(y^2\):

Working
\[ 2x+2y\frac{dy}{dx}=0\ \Longrightarrow\ \frac{dy}{dx}=-\frac{x}{y} \]
Review

Chapter Summary

Definition

\(f'(x)=\lim_{h\to0}\tfrac{f(x+h)-f(x)}{h}\) — slope of the tangent.

Differentiability

Needs \(\text{LHD}=\text{RHD}\); implies continuity, not conversely.

Standard table

\((x^n)'=nx^{n-1}\), \((\sin x)'=\cos x\), \((e^x)'=e^x\), \((\ln x)'=\tfrac1x\).

Combining rules

Product \(f'g+fg'\), quotient \(\tfrac{f'g-fg'}{g^2}\), chain \(f'(g)\,g'\).

Special techniques

Implicit, logarithmic, parametric \(\tfrac{dy/dt}{dx/dt}\), inverse \(\tfrac{1}{dx/dy}\).

Higher order

\(f''=\tfrac{d}{dx}(f')\); inverse-trig derivatives from the inverse rule.

Practice

Problems

Name the structure first — power, product, quotient, composite, implicit — then apply the matching rule. Difficulty rises down the list.

  1. Differentiate \(y=x^5-3x^2+7\).
  2. Differentiate \(y=\sqrt{x}+\dfrac1x\).
  3. Differentiate \(y=x^3e^x\).
  4. Differentiate \(y=\dfrac{\sin x}{x}\).
  5. Differentiate \(y=\cos(5x^3)\).
  6. Differentiate \(y=\ln(x^2+1)\).
  7. Differentiate \(y=e^{\sqrt{x}}\).
  8. If \(y=x^{\sin x}\), find \(\dfrac{dy}{dx}\).
  9. For the curve \(x=t^2,\ y=t^3\), find \(\dfrac{dy}{dx}\).
  10. Find \(\dfrac{dy}{dx}\) for \(4x^2+9y^2=36\).
  11. Differentiate \(y=\tan^{-1}x+\sin^{-1}x\).
  12. Find the second derivative of \(y=x^4-3x^3+2x\).
Tip: identify the outermost operation before differentiating — is the expression fundamentally a product, a quotient, or a composite? That decides the first rule to apply, and the chain rule then handles whatever sits inside. For powers with a variable in both base and exponent, such as \(x^x\) or \(x^{\sin x}\), reach straight for logarithmic differentiation.