Derivatives

The derivative $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$ is the instantaneous rate of change — the slope of the tangent. Everything below is machinery for computing it without touching the limit.

Power rule and linearity

$\frac{d}{dx}\,x^n = n\,x^{n-1} \qquad (cf)' = c\,f' \qquad (f+g)' = f' + g'$

Works for all real $n$ — rewrite roots and fractions as powers first.

Worked example. $f(x) = 4x^3 - \dfrac{2}{x^2} + 3\sqrt{x}$

Rewrite: $f(x) = 4x^3 - 2x^{-2} + 3x^{1/2}$

$f'(x) = 12x^2 + 4x^{-3} + \tfrac{3}{2}x^{-1/2} = 12x^2 + \frac{4}{x^3} + \frac{3}{2\sqrt{x}}$

Exercise 1.1

Differentiate by hand, writing each function as powers first:

a) $f(x) = 7x^5 - 3x^2 + 12$

b) $g(x) = \dfrac{5}{x^3} + \sqrt[3]{x}$

c) $h(x) = \dfrac{x^2 + 1}{\sqrt{x}}$ (split the fraction first)

Product rule

$(fg)' = f'g + fg'$

Worked example. $y = x^2 e^x$

$y' = 2x\,e^x + x^2 e^x = x e^x (2 + x)$

Recall the standard derivatives you'll combine with it: $(e^x)' = e^x$, $(\ln x)' = \frac1x$, $(\sin x)' = \cos x$, $(\cos x)' = -\sin x$.

Where these come from

$e^x$ — for any base, $(a^x)' = a^x \ln a$: the exponential reproduces itself up to a constant factor. $e \approx 2.718$ is defined as the base where that factor is $1$, i.e. the slope of $e^x$ equals its value everywhere. That self-reproduction is why $e^x$ shows up in every growth/decay ODE: $y' = ky \Rightarrow y = Ce^{kx}$.

$\ln x$ — inverse of $e^x$. Differentiate $e^{\ln x} = x$ implicitly: $e^{\ln x} \cdot (\ln x)' = 1$, so $(\ln x)' = 1/e^{\ln x} = 1/x$. Same trick gives $(\log_a x)' = \frac{1}{x \ln a}$.

$\sin, \cos$ — differentiation is a quarter-turn phase shift: $\sin x \to \cos x \to -\sin x \to -\cos x \to \sin x$, a cycle of four (equivalently $(\sin x)' = \sin(x + \tfrac{\pi}{2})$). That's why the second derivative of each is its own negative — the source of $y'' = -\omega^2 y$ in every oscillator problem.

Exercise 1.2

Differentiate and simplify:

a) $f(x) = x^3 \ln x$

b) $g(x) = e^x \sin x$

c) $h(t) = (t^2 + 1)(t^3 - 2t)$ — twice: once with the product rule, once by expanding first. Confirm both agree.

Quotient rule

$\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}$

Numerator order matters — "low d-high minus high d-low".

Worked example. $y = \dfrac{\sin x}{x}$

$y' = \frac{x\cos x - \sin x}{x^2}$

Exercise 1.3

a) $f(x) = \dfrac{x^2}{x + 1}$

b) $g(x) = \dfrac{e^x}{x^2 + 1}$

c) $\tan x = \dfrac{\sin x}{\cos x}$ — derive $(\tan x)' = \dfrac{1}{\cos^2 x}$ yourself.

Chain rule

$\frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x)$

Differentiate the outer function at the inner one, times the derivative of the inner. For nested compositions, work outside-in.

Worked example. $y = \sin(x^2)$: outer $\sin$, inner $x^2$:

$y' = \cos(x^2) \cdot 2x$

Worked example (double chain). $y = e^{\cos(3x)}$:

$y' = e^{\cos(3x)} \cdot (-\sin(3x)) \cdot 3 = -3\sin(3x)\,e^{\cos(3x)}$

Exercise 1.4

a) $f(x) = (2x^3 - 5)^7$

b) $g(x) = \ln(x^2 + 4)$

c) $h(x) = \sqrt{1 + e^{2x}}$

d) $k(x) = x^2 \sin(3x)$ (product and chain)

See it

A derivative is a slope. Change f and a below; the tangent at $x = a$ uses the slope from the rules above.

import numpy as np
import matplotlib.pyplot as plt

f = lambda x: x**3 - 2*x       # try your own function
df = lambda x: 3*x**2 - 2      # ... and its derivative
a = 1.0                        # point of tangency

x = np.linspace(-2.5, 2.5, 400)
tangent = f(a) + df(a) * (x - a)

plt.plot(x, f(x), label='f(x)')
plt.plot(x, tangent, '--', label=f'tangent at x={a}, slope={df(a):.2f}')
plt.scatter([a], [f(a)], zorder=3)
plt.ylim(-6, 6)
plt.legend()
plt.grid(True)
plt.show()

Mixed practice

No scaffolding — pick the right rule(s) yourself.

a) $y = \dfrac{\ln x}{x^2}$

b) $y = e^{-x^2/2}$ (the Gaussian — you will meet it weekly in physics)

c) $y = x\,e^x \cos x$ (three factors)

d) $y = \left(\dfrac{x}{1+x}\right)^3$