IMOClass 11 › Derivatives

Derivatives

Derivative from First Principles

The derivative of a function measures its instantaneous rate of change — the slope of the tangent to the curve at a point. It is defined as a limit of average rates of change, known as the first principle (or $ab\,initio$) definition:

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

The quotient $\dfrac{f(x+h) - f(x)}{h}$ is the slope of the secant joining the points $(x,\, f(x))$ and $(x+h,\, f(x+h))$. As $h \to 0$ the second point slides towards the first, and the secant pivots into the tangent; its slope is the derivative.

Derivatives from first principles. Applying the definition to the basic functions gives results worth knowing on sight:

Function $f(x)$Derivative $f'(x)$
$k$ (constant)$0$
$x^n$$n\,x^{n-1}$
$\sin x$$\cos x$
$\cos x$$-\sin x$
$e^x$$e^x$
$\ln x$$\dfrac{1}{x}$

Rules of differentiation. Once the basics are known, four rules let you combine them. For differentiable functions $u$ and $v$:

$$(u \pm v)' = u' \pm v' \qquad (k\,u)' = k\,u'$$
$$\textbf{Product rule: } (uv)' = u'\,v + u\,v'$$
$$\textbf{Quotient rule: } \left(\dfrac{u}{v}\right)' = \dfrac{u'\,v - u\,v'}{v^{2}}, \;\; v \ne 0$$

A frequent slip is to assume $(uv)' = u'v'$ — it is not. The product rule keeps both cross terms, and the quotient rule subtracts (order matters: it is $u'v - uv'$ on top, never the reverse).

Deeper Insight — the derivative is a limit of slopes: Everything in this topic rests on one picture. The fraction $\dfrac{f(x+h)-f(x)}{h}$ is just "rise over run" between two points on the curve — an average rate of change across a step of width $h$. Shrinking $h$ towards zero is what converts that average into an instantaneous rate, and geometrically it is the secant line rotating until it rests as the tangent. This is also why first principles repeatedly throw up the standard limits of the previous topic: differentiating $\sin x$ forces $\lim_{h\to 0}\tfrac{\sin h}{h}=1$ to surface, and differentiating $e^x$ forces $\lim_{h\to 0}\tfrac{e^h-1}{h}=1$. The table of derivatives and the four rules are simply the compiled, reusable output of this limit process; understanding that they come from a limit — rather than appearing by decree — is what makes the whole of differential calculus cohere.

Secant line approaching the tangent as h shrinks to zero Secant → Tangent as h → 0 x x + h secant (slope = average rate) tangent slope of secant → f ′(x) as the gap h → 0 Product and quotient rule structure Two rules to keep straight PRODUCT(uv)′ = u′v + uv′ QUOTIENT(u/v)′ = (u′v − uv′)/v²
Example 1: Find the derivative of $f(x) = x^2$ from first principles.
  1. Apply the definition: $f'(x) = \lim_{h \to 0}\dfrac{(x+h)^2 - x^2}{h}$.
  2. Expand the numerator: $(x+h)^2 - x^2 = x^2 + 2xh + h^2 - x^2 = 2xh + h^2$.
  3. $f'(x) = \lim_{h \to 0}\dfrac{2xh + h^2}{h} = \lim_{h \to 0}(2x + h)$.
  4. Let $h \to 0$: $f'(x) = 2x$.

Answer: $f'(x) = 2x$.

Example 2: Find the derivative of $f(x) = \sin x$ from first principles.
  1. $f'(x) = \lim_{h \to 0}\dfrac{\sin(x+h) - \sin x}{h}$.
  2. Use $\sin(x+h) - \sin x = 2\cos\!\left(x + \dfrac{h}{2}\right)\sin\dfrac{h}{2}$.
  3. $f'(x) = \lim_{h \to 0}\cos\!\left(x + \dfrac{h}{2}\right)\cdot\dfrac{\sin(h/2)}{h/2}$.
  4. As $h \to 0$: $\dfrac{\sin(h/2)}{h/2} \to 1$ and $\cos\!\left(x+\dfrac{h}{2}\right) \to \cos x$.

Answer: $f'(x) = \cos x$.

Example 3: Differentiate $y = 4x^3 - 7x^2 + 5x - 9$.
  1. Differentiate term by term using $(x^n)' = n\,x^{n-1}$ and the constant-multiple rule.
  2. $\dfrac{d}{dx}(4x^3) = 12x^2$, $\dfrac{d}{dx}(-7x^2) = -14x$, $\dfrac{d}{dx}(5x) = 5$, $\dfrac{d}{dx}(-9) = 0$.
  3. Add: $\dfrac{dy}{dx} = 12x^2 - 14x + 5$.

Answer: $\dfrac{dy}{dx} = 12x^2 - 14x + 5$.

Example 4: Differentiate $y = x^2 \sin x$ using the product rule.
  1. Let $u = x^2$ and $v = \sin x$; then $u' = 2x$ and $v' = \cos x$.
  2. Product rule: $\dfrac{dy}{dx} = u'v + uv' = 2x\sin x + x^2\cos x$.

Answer: $\dfrac{dy}{dx} = 2x\sin x + x^2\cos x$.

Example 5: Differentiate $y = \dfrac{\sin x}{x}$ using the quotient rule.
  1. Let $u = \sin x$ and $v = x$; then $u' = \cos x$ and $v' = 1$.
  2. Quotient rule: $\dfrac{dy}{dx} = \dfrac{u'v - uv'}{v^2} = \dfrac{x\cos x - \sin x}{x^2}$.

Answer: $\dfrac{dy}{dx} = \dfrac{x\cos x - \sin x}{x^2}$.

Example 6: Find the slope of the tangent to $y = x^3$ at the point $x = 2$.
  1. The slope of the tangent is the derivative evaluated at the point.
  2. $\dfrac{dy}{dx} = 3x^2$.
  3. At $x = 2$: $3(2)^2 = 3 \times 4 = 12$.

Answer: The slope of the tangent at $x = 2$ is $12$.

Quick recap
  • First principle: $f'(x) = \lim_{h \to 0}\dfrac{f(x+h) - f(x)}{h}$ — the limit of the secant slope.
  • Key derivatives: $(x^n)' = n\,x^{n-1}$, $(\sin x)' = \cos x$, $(\cos x)' = -\sin x$.
  • Product rule: $(uv)' = u'v + uv'$ — never $u'v'$.
  • Quotient rule: $\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}$ — order on top matters.
  • The derivative at a point is the slope of the tangent (instantaneous rate of change) there.
✓ Quick check
The derivative of eˣ+ln x is:
Differentiate both terms separately.
The derivative of 5cos x is:
d/dx(cos x)=−sin x.

Rules of Differentiation

Power rule d/dx(xⁿ) = nxⁿ⁻¹. Product: (uv)′ = u′v + uv′. Quotient: (u/v)′ = (u′v − uv′)/v².

Example 1: Differentiate x⁵.
5x⁴.
Example 2: Differentiate x²·sinx (product rule).
2x sinx + x² cosx.
Quick recap
  • d/dx(xⁿ) = nxⁿ⁻¹.
  • Use product and quotient rules for combinations.
✓ Quick check
If y=1/x, then dy/dx equals:
y=x⁻¹, so derivative is −x⁻²=−1/x².
If u and v are two differentiable functions of x (where v ≠ 0), the quotient rule states that d/dx (u/v) is:
The quotient rule is d/dx (u/v) = (v du/dx − u dv/dx) / v².

Derivatives of Standard Functions

Know d/dx: sinx = cosx, cosx = −sinx, eˣ = eˣ, ln x = 1/x. Combine with the rules above.

Example 1: Differentiate sinx + cosx.
cosx − sinx.
Example 2: Derivative of 3eˣ.
3eˣ.
Quick recap
  • (sinx)′ = cosx, (cosx)′ = −sinx.
  • (eˣ)′ = eˣ, (ln x)′ = 1/x.
✓ Quick check
The velocity of a Delhi Metro train during its journey between two stations is modeled by v(t) = 20t − t² (m/s). What is the acceleration of the train at t = 5 seconds?
Acceleration is the derivative of velocity, a(t) = dv/dt = 20 − 2t. At t = 5, a(5) = 20 − 10 = 10 m/s².
A bacteria culture's population is given by P(t) = 1000 + 50t². What is the growth rate of the population at t = 4 hours?
Growth rate dP/dt = 100t. At t = 4, dP/dt = 100(4) = 400.
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