在線即時計算器 · 純前端本地秒算 · 零數據上傳

∫ Tanh First Derivative 計算機

專業免費的 Tanh First Derivative 計算機。純前端本機毫秒級計算、零資料上傳、極致隱私保護。提供詳細公式推導、實例示範與常見問答。

即時推算結果 (Computed Output)
0.786448 = 1 - tanh(x)^2
專業公認計算模型,即時運算完畢
📜 計算歷史記錄 (試算軌跡)
尚無歷史記錄,每次點擊「計算」將在此留下試算軌跡...

What this calculator does

This tool evaluates the hyperbolic tangent function \tanh(x) and, more importantly, its first derivative \tanh'(x) at any real input x. It also reports the second derivative \tanh''(x). The hyperbolic tangent is a smooth, S-shaped (sigmoidal) function whose output is bounded between -1 and 1, which is why it appears so often as a neural-network activation function and in physics and engineering models.

How to use it

Enter any real number for x and submit. The calculator computes \tanh(x) once, then derives both the first and second derivatives from that single value. There are no units to convert: x is a dimensionless real number, and all outputs are dimensionless as well.

The formula explained

The hyperbolic tangent is

\tanh(x) = ((e^x - e^{-x}) ÷ (e^x + e^{-x})).
Its first derivative has the elegant closed form
f'(x) = 1 - \tanh^{2}(x),
equivalently sech^2(x) = ((1) ÷ (\cosh^2(x))). Because \tanh(x) lies in (-1, 1), the first derivative always lies in (0, 1], peaking at x = 0 where the slope is exactly 1. The second derivative is
f''(x) = -2\tanh(x)≤ft(1 - \tanh^{2}(x)\right),
an odd function that crosses zero at x = 0.

Worked example (x = 0.5)

\tanh(0.5) = ((1.6487212707 - 0.6065306597) ÷ (1.6487212707 + 0.6065306597)) = 0.4621171573.
Then
f'(0.5) = 1 - 0.4621171573^{2} = 0.7864477541,
and
f''(0.5) = -2 × 0.4621171573 × 0.7864477541 = -0.7269278407.

使用注意事項與精準校對原則(6 大核心守則)

📐 1. 嚴格對齊輸入量綱與單位

輸入各項數值時請務必確認量綱(例如小時、天數、金額、百分比),避免單位不匹配引發推算偏差。

🔒 2. 純前端本機毫秒級隱私保障

所有計算邏輯 100% 在您的終端瀏覽器內完成,資料絕不上傳伺服器,嚴格守護商業與個人隱私。

📊 3. 符合公認學術與產業計算模型

演算法嚴格遵循國際通用數學公式與標準計算模型,經多重情境交叉核對,數值嚴密精確。

📋 4. 一鍵複製與免登入 PDF 匯出

支援一鍵複製推算結果字串至剪貼簿,亦可直接點擊列印按鈕輸出純淨 PDF 工作報表。

🔄 5. 多情景敏感度動態對比

建議針對極端值與多組預期情景分別試算,觀察輸出數據的波動區間,輔助做出更穩健的決策。

🛡️ 6. 專業諮詢與日常參考免責聲明

本工具運算結果供日常排程、學業研習與事前參考之用,如涉及重大法規或投資決策請諮詢持證專家。

常見問題 (FAQ)

Why does the gradient vanish for large x?
As \(x\) grows, tanh saturates toward \(\pm 1\), so \(1 - \tanh^2\) approaches 0. This "vanishing gradient" can slow training in deep networks.
Is the derivative ever negative?
No. \(f'(x) = \operatorname{sech}^2(x)\) is strictly positive for all real \(x\), so tanh is always increasing.
Is there any divide-by-zero risk?
No. \(\cosh(x)\) is at least 1 for every real \(x\), so \(\operatorname{sech}^2(x)\) is always well defined.