Derivatives

100. Derivatives

Measures the instantaneous rate of change of a function

Derivative of of a function $𝑓 (𝑥)$

𝑓^{'} (𝑥) = \frac{𝑑}{𝑑 𝑥} 𝑓 (𝑥)

General Pattern:

Linear term: If a function is linear (e.g., $𝑎 𝑥 + 𝑏$ ), the first derivative will be a constant $𝑎$ .
Constant term: The derivative of a constant is zero.
Higher-degree polynomials: For polynomials of higher degrees, each derivative decreases the degree of the polynomial until you get a constant, after which all further derivatives are zero.

Example

Consider the quartic function: $𝑓 (𝑥) = 𝑥^{4} - 2 𝑥^{2} + 1$

First derivative: $𝑓^{'} (𝑥) = 4 𝑥^{3} - 4 𝑥$ (cubic)
Second derivative: $𝑓^{″} (𝑥) = 12 𝑥^{2} - 4$ (quadratic)
Third derivative: $𝑓^{‴} (𝑥) = 24 𝑥$ (linear)
Fourth derivative: $𝑓^{⁗} (𝑥) = 24$ (constant)
Fourth derivative: $𝑓^{'''''} (𝑥) = 0$ (zero)

100.1. Power Rule

\begin{aligned} 𝑓 (𝑥) & = 𝑥^{𝑛} \\ 𝑓^{'} (𝑥) & = 𝑛 𝑥^{𝑛 - 1} \end{aligned}

Example

\begin{aligned} 𝑓 (𝑥) & = 𝑥^{3} \\ 𝑓^{'} (𝑥) & = 3 𝑥^{2} \end{aligned}

100.2. Product Rule

\begin{aligned} 𝑓 (𝑥) & = 𝑢 (𝑥) \cdot 𝑣 (𝑥) \\ 𝑓^{'} (𝑥) & = 𝑢^{'} (𝑥) 𝑣 (𝑥) + 𝑢 (𝑥) 𝑣^{'} (𝑥) \end{aligned}

Example

𝑓 (𝑥) = 𝑥^{2} \cdot 𝑥^{3}

𝑢 (𝑥) = 𝑥^{2}

𝑣 (𝑥) = 𝑥^{3}

Compute the derivatives of $𝑢 (𝑥)$ and $𝑣 (𝑥)$

𝑢^{'} (𝑥) = 2 𝑥

𝑣^{'} (𝑥) = 3 𝑥^{2}

Apply rule

\begin{aligned} 𝑓^{'} (𝑥) & = (2 𝑥) (𝑥^{3}) + (𝑥^{2}) (3 𝑥^{2}) \\ = 2 𝑥^{2} + 3 𝑥^{4} \\ = 5 𝑥^{4} \end{aligned}

100.3. Quotient Rule

\begin{aligned} 𝑓 (𝑥) & = \frac{𝑢 (𝑥)}{𝑣 (𝑥)} \\ 𝑓^{'} (𝑥) & = \frac{𝑢^{'} (𝑥) 𝑣 (𝑥) - 𝑢 (𝑥) 𝑣^{'} (𝑥)}{{𝑣 (𝑥)}^{2}} \end{aligned}

Example

𝑓 (𝑥) = \frac{𝑥^{3}}{𝑥^{2}}

𝑢 (𝑥) = 𝑥^{3}

𝑣 (𝑥) = 𝑥^{2}

Compute the derivatives of $𝑢 (𝑥)$ and $𝑣 (𝑥)$

𝑢^{'} (𝑥) = 3 𝑥^{2}

𝑣^{'} (𝑥) = 2 𝑥

Apply rule

\begin{aligned} 𝑓^{'} (𝑥) & = \frac{(3 𝑥^{2}) (𝑥^{2}) - (𝑥^{3}) (2 𝑥)}{{(𝑥^{2})}^{2}} \\ = \frac{3 𝑥^{4} - 2 𝑥^{4}}{𝑥^{4}} \\ = \frac{𝑥^{4}}{𝑥^{4}} \\ = 1 \end{aligned}

100.4. Chain Rule

\begin{matrix} 𝑓 (𝑔 (𝑥)) \\ 𝑓^{'} (𝑥) = 𝑓^{'} (𝑔 (𝑥)) \cdot 𝑔^{'} (𝑥) \end{matrix}

Example

𝑓 (𝑥) = {(2 𝑥 + 1)}^{3}

Outer function:

𝑓 (𝑢) = 𝑢^{3}

Inner function

𝑔 (𝑥) = 2 𝑥 + 1

Compute the derivatives of $𝑓^{'} (𝑢)$ and $𝑔^{'} (𝑥)$

𝑓^{'} (𝑢) = 3 𝑢^{2}

𝑔^{'} (𝑥) = 2

Apply rule

\begin{aligned} 𝑓^{'} (𝑥) & = 3 {(2 𝑥 + 1)}^{2} \cdot 2 \\ = 6 {(2 𝑥 + 1)}^{2} \end{aligned}

100.5. Others

1. Trigonometric Functions

\frac{𝑑}{𝑑 𝑥} [sin (𝑥)] = cos (𝑥)

\frac{𝑑}{𝑑 𝑥} [cos (𝑥)] = - sin (𝑥)

\frac{𝑑}{𝑑 𝑥} [tan (𝑥)] = {sec}^{2} (𝑥)

\frac{𝑑}{𝑑 𝑥} [cot (𝑥)] = {csc}^{2} (𝑥)

\frac{𝑑}{𝑑 𝑥} [sec (𝑥)] = sec (𝑥) tan (𝑥)

\frac{𝑑}{𝑑 𝑥} [csc (𝑥)] = - csc (𝑥) cot (𝑥)

2. Inverse Trigonometric Functions

\frac{𝑑}{𝑑 𝑥} [arcsin (𝑥)] = \frac{1}{\sqrt{1 - 𝑥^{2}}}

\frac{𝑑}{𝑑 𝑥} [arccos (𝑥)] = - \frac{1}{\sqrt{1 - 𝑥^{2}}}

\frac{𝑑}{𝑑 𝑥} [arctan (𝑥)] = \frac{1}{1 + 𝑥^{2}}

\frac{𝑑}{𝑑 𝑥} [arccot (𝑥)] = - \frac{1}{1 + 𝑥^{2}}

\frac{𝑑}{𝑑 𝑥} [arcsec (𝑥)] = \frac{1}{| 𝑥 | \sqrt{𝑥^{2} - 1}}

\frac{𝑑}{𝑑 𝑥} [arccsc (𝑥)] = - \frac{1}{| 𝑥 | \sqrt{𝑥^{2} - 1}}

3. Exponential and Logarithmic Functions

\frac{𝑑}{𝑑 𝑥} [𝑒^{𝑥}] = 𝑒^{𝑥}

\frac{𝑑}{𝑑 𝑥} [𝑎^{𝑥}] = 𝑎^{𝑥} ln (𝑎)

\frac{𝑑}{𝑑 𝑥} [ln (𝑥)] = \frac{1}{𝑥}

\frac{𝑑}{𝑑 𝑥} [{log}_{𝑎 (𝑥)}] = \frac{1}{𝑥 ln (𝑎)}

Definition 0: Stationary Point

A stationary point of a function $𝑓 (𝑥)$ is a solution to

𝑓' (𝑥) = 0 That is, where the slope (derivative) is zero. In 1D, this means the tangent line is horizontal In nD, it means the gradient vector \nabla 𝑓 (𝑥) = 0← Limits Higher-Order Derivatives →window.addEventListener('DOMContentLoaded', () => { if (typeof PagefindUI === 'undefined') return; const addTextFrag = (url, q) => {   if (!q) return url;   const enc = encodeURIComponent(q);   if (url.includes(':~:text=')) return url;   return url.includes('#') ? url + ':~:text=' + enc                            : url + '#:~:text=' + enc; }; new PagefindUI({   element: '#search',   showSubResults: true,   processResult: (result) => {     const inp = document.querySelector('.pagefind-ui__search-input');     const q = inp ? inp.value.trim() : '';     result.url = addTextFrag(result.url, q);     if (result.sub_results) {       result.sub_results.forEach(sr => { sr.url = addTextFrag(sr.url, q); });     }     return result;   }, });});(() => { const root = document.documentElement, body = document.body; const W_KEY = 'ieor-sidebar-w', C_KEY = 'ieor-sidebar-collapsed'; const MOBILE = 900; const savedW = localStorage.getItem(W_KEY); if (savedW) root.style.setProperty('--sidebar-w', savedW + 'px'); if (localStorage.getItem(C_KEY) === '1' && window.innerWidth > MOBILE)   body.classList.add('sidebar-collapsed'); const btn = document.querySelector('.nav-toggle'); if (btn) btn.addEventListener('click', () => {   if (window.innerWidth <= MOBILE) {     body.classList.toggle('nav-open');   } else {     body.classList.toggle('sidebar-collapsed');     localStorage.setItem(C_KEY, body.classList.contains('sidebar-collapsed') ? '1' : '0');   } }); const rz = document.querySelector('.sidebar-resizer'); if (rz) rz.addEventListener('pointerdown', (e) => {   e.preventDefault();   rz.setPointerCapture(e.pointerId);   body.classList.add('resizing');   const COLLAPSE_AT = 120;   let cleanup;   const onMove = (ev) => {     if (ev.clientX < COLLAPSE_AT) {       body.classList.add('sidebar-collapsed');       localStorage.setItem(C_KEY, '1');       cleanup();       return;     }     const w = Math.min(600, Math.max(180, ev.clientX));     root.style.setProperty('--sidebar-w', w + 'px');   };   cleanup = () => {     try { rz.releasePointerCapture(e.pointerId); } catch (_) {}     body.classList.remove('resizing');     rz.removeEventListener('pointermove', onMove);     rz.removeEventListener('pointerup', cleanup);     const w = parseInt(getComputedStyle(root).getPropertyValue('--sidebar-w'));     if (!isNaN(w)) localStorage.setItem(W_KEY, String(w));   };   rz.addEventListener('pointermove', onMove);   rz.addEventListener('pointerup', cleanup); });})();