Lecture 3: Differentiation PDF

LECTURE 3: DIFFERENTIATION 𝒇(𝒙) 𝑑𝑓 𝑓(𝑥 + ℎ) − 𝑓(𝑥) = 𝑓 (𝑥) = lim 𝑑𝑥 → ℎ may be interpreted as a slope of the tangent lin...

LECTURE 3: DIFFERENTIATION 𝒇(𝒙) 𝑑𝑓 𝑓(𝑥 + ℎ) − 𝑓(𝑥) = 𝑓 (𝑥) = lim 𝑑𝑥 → ℎ may be interpreted as a slope of the tangent line to the graph of 𝑓(𝑥) at the point (𝑥, 𝑓(𝑥)) DIFFERENTIATION RULES 𝒇(𝒙) 𝒇(𝒙) = 𝒄 𝒅𝒇 𝒅 = (𝒄) = 𝟎 𝒅𝒙 𝒅𝒙 𝒅 𝒏 (𝒙 ) = 𝒏 𝒙𝒏 𝟏 𝒅𝒙 𝒇(𝒙) 𝒙 𝒄 𝒅 𝒅𝒇 (𝒄 𝒇 ) = 𝒄 𝒅𝒙 𝒅𝒙 𝒇(𝒙) 𝒈(𝒙) 𝒅 𝒅 𝒅 [𝒇(𝒙) ± 𝒈(𝒙)] = 𝒇(𝒙) ± 𝒈(𝒙) 𝒅𝒙 𝒅𝒙 𝒅𝒙 33  𝒇(𝒙) 𝒈(𝒙) 𝑑 𝑑 𝑑 [𝑓 (𝑥 )𝑔(𝑥)] = 𝑓 (𝑥 ) 𝑔(𝑥 ) + 𝑔(𝑥 ) 𝑓 (𝑥 ) 𝑑𝑥 𝑑𝑥 𝑑𝑥 If 𝑓(𝑥) and 𝑔(𝑥) are both differentiable, then 𝑑 𝑑 𝑑 𝑓 (𝑥 ) 𝑔 ( 𝑥 ) 𝑓 ( 𝑥 ) − 𝑓 (𝑥 ) 𝑔 ( 𝑥 ) = 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑔(𝑥 ) [𝑔(𝑥 )] Derivatives of Elementary 𝒇(𝒙) 𝒅𝒇 𝒅𝒙 𝒙𝒏 𝒏 𝒙𝒏 𝟏 𝒂𝒙 , 𝒂 > 𝟎, 𝒂 ≠ 𝟏 𝒂𝒙 ∗ 𝐥𝐧 𝒂 𝒆𝒙 𝒆𝒙 𝐥𝐨𝐠 𝒂 𝒙 , 𝒂 > 𝟎, 𝒂 ≠ 𝟏 𝟏 , 𝒙>𝟎 𝒙 𝐥𝐧 𝒂 𝐥𝐧 𝒙 𝟏 , 𝒙>𝟎 𝒙 𝐬𝐢𝐧 𝒖 , 𝒖 = 𝒇(𝒙) 𝒅𝒖 𝐜𝐨𝐬 𝒖 ∗ 𝒅𝒙 𝐜𝐨𝐬 𝒖 𝒅𝒖 − 𝐬𝐢𝐧 𝒖 ∗ 𝒅𝒙 34  𝐭𝐚𝐧 𝒖 𝒅𝒖 𝐬𝐞𝐜 𝟐 𝒖 ∗ 𝒅𝒙 𝐬𝐞𝐜 𝒖 𝒅𝒖 𝐬𝐞𝐜 𝒖 ∗ 𝐭𝐚𝐧 𝒖 ∗ 𝒅𝒙 𝐜𝐬𝐜 𝒖 𝒅𝒖 −𝐜𝐬𝐜 𝒖 ∗ 𝐜𝐨𝐭 𝒖 ∗ 𝒅𝒙 𝐜𝐨𝐭 𝒖 𝒅𝒖 − 𝒄𝒔𝒄𝟐 𝒖 ∗ 𝒅𝒙 Inverse Trigonometric Functions 𝐬𝐢𝐧 𝟏 𝒖 𝟏 ∗𝒖 |𝒖| < 𝟏 √𝟏 − 𝒖𝟐 𝐜𝐨𝐬 𝟏 𝒖 −𝟏 ∗𝒖 |𝒖| < 𝟏 √𝟏 − 𝒖 𝟐 𝐭𝐚𝐧 𝒖𝟏 𝟏 ∗𝒖 𝟏 + 𝒖𝟐 𝐬𝐞𝐜 𝟏 𝒖 𝟏 ∗𝒖 |𝒖| > 𝟏 𝒖√𝒖𝟐 − 𝟏 𝐜𝐬𝐜 𝟏 𝒖 −𝟏 ∗𝒖 |𝒖| > 𝟏 𝟐 𝒖√𝒖 − 𝟏 𝐜𝐨𝐭 𝒖𝟏 −𝟏 ∗𝒖 𝟏 + 𝒖𝟐 Hyperbolic Functions 𝐬𝐢𝐧𝐡 𝒖 𝐜𝐨𝐬𝐡 𝒖 ∗ 𝒖 𝐜𝐨𝐬𝐡 𝒖 𝐬𝐢𝐧𝐡 𝒖 ∗ 𝒖 𝐭𝐚𝐧𝐡 𝒖 𝐬𝐞𝐜𝐡𝟐 𝒖 ∗ 𝒖 𝐬𝐞𝐜𝐡 𝒖 − 𝐬𝐞𝐜𝐡 𝒖 ∗ 𝐭𝐚𝐧𝐡 𝒖 ∗ 𝒖 𝐜𝐬𝐜𝐡 𝒖 −𝐜𝐬𝐜𝐡 𝒖 ∗ 𝐜𝐨𝐭𝐡 𝒖 ∗ 𝒖 𝐜𝐨𝐭𝐡 𝒖 − 𝒄𝒔𝒄𝒉𝟐 𝒖 ∗ 𝒖 35  Inverse Hyperbolic Functions 𝐬𝐢𝐧𝐡 𝟏 𝒖 𝟏 ∗𝒖 √𝟏 + 𝒖𝟐 𝟏 𝐜𝐨𝐬𝐡 𝒖 𝟏 ∗𝒖 |𝒖| > 𝟏 √𝒖𝟐 −𝟏 𝐭𝐚𝐧𝐡 𝟏 𝒖 𝟏 ∗𝒖 |𝒖| < 𝟏 𝟏 − 𝒖𝟐 𝐬𝐞𝐜𝐡 𝟏 𝒖 −𝟏 ∗𝒖 |𝒖| < 𝟏 𝒖√𝟏 − 𝒖𝟐 𝐜𝐬𝐜𝐡 𝟏 𝒖 −𝟏 ∗𝒖 𝒖√𝟏 + 𝒖𝟐 𝐜𝐨𝐭𝐡 𝟏 𝒖 −𝟏 ∗𝒖 |𝒖 | > 𝟏 𝒖𝟐 − 𝟏 𝒚 𝟓 𝟐 𝟏 𝟏) 𝒚 = 𝟐𝒙𝟑 + √𝒙 + 𝒙𝟑 + + 𝒙𝟑 𝟒 √𝒙 𝟏 𝟑 𝟏 𝟑 𝟑 𝒚 = 𝟐𝒙 + 𝒙𝟐 + 𝒙𝟓 + 𝟐𝒙 + 𝒙𝟒 𝟏 𝟐 𝟓 𝑦 = 𝟔𝒙𝟐 + 𝟎. 𝟓𝒙 𝟐 + 𝟎. 𝟔𝒙 𝟓 − 𝟔𝒙 𝟒 − 𝟎. 𝟐𝟓𝒙 𝟒 𝟐) 𝒚 = 𝒔𝒊𝒏 𝟓𝒙 + 𝒔𝒆𝒄 𝟐𝒙 + 𝟒 𝐥𝐧(𝒙𝟐 + 𝟏) + 𝟐𝒆𝟑𝒙 𝟓 𝟒 ∗ 𝟐𝒙 𝒚 = 𝟓 𝒄𝒐𝒔 𝟓𝒙 + 𝟐 𝒔𝒆𝒄 𝟐𝒙 𝒕𝒂𝒏 𝟐𝒙 + 𝟐 + 𝟔 𝒆𝟑𝒙 𝟓 𝒙 +𝟏 𝟑) 𝒚 = 𝒙𝟐 + 𝟏 + 𝐭𝐚𝐧(𝒙𝟐 + 𝟓𝒙) + 𝐜𝐨𝐬𝐡(𝒙𝟐 + 𝟏) 𝒚 = 𝟎. 𝟓(𝒙𝟐 + 𝟏) 𝟎.𝟓 ∗ 𝟐𝒙 + (𝟐𝒙 + 𝟓) 𝒔𝒆𝒄𝟐 (𝒙𝟐 + 𝟓𝒙) + 𝟐𝒙 𝐬𝐢𝐧𝐡(𝒙𝟐 + 𝟏) 36 𝟒) 𝒚 = 𝒙𝟑 𝒔𝒊𝒏𝟐 (𝟐𝒙) + 𝒆𝒄𝒐𝒔𝒉 𝒙 + 𝐜𝐨𝐬 √𝒙 𝒚 = 𝟑𝒙𝟐 𝒔𝒊𝒏𝟐 (𝟐𝒙) + 𝟒𝒙𝟑 𝐬𝐢 (𝟐𝒙) 𝐜𝐨 𝐬(𝟐𝒙) + 𝒆𝒄𝒐𝒔𝒉 𝒙 ∗ 𝐬𝐢𝐧 𝟏 − 𝐬𝐢𝐧 √𝒙 𝟐 √𝒙 𝒔𝒆𝒄 𝒙𝟑 𝟓) 𝒚 = 𝟏 + 𝒕𝒂𝒏 𝒙𝟓 𝟏 + 𝒕𝒂𝒏 𝒙𝟓 ∗ 𝟑𝒙𝟐 ∗ 𝒔𝒆𝒄 𝒙𝟑 𝒕𝒂𝒏 𝒙𝟑 − 𝒔𝒆𝒄 𝒙𝟑 ∗ 𝟓𝒙𝟒 ∗ 𝒔𝒆𝒄𝟐 𝒙𝟓 𝒚 = (𝟏 + 𝒕𝒂𝒏 𝒙𝟓 )𝟐 𝟔) 𝒚 = 𝒙𝟐 ∗ 𝒔𝒊𝒏(𝒙𝟒 + 𝟓) ∗ 𝐭𝐚𝐧√𝒙 + 𝟐 + 𝐜𝐨𝐭 √𝒙 𝒚 = 𝟐 ∗ 𝒔𝒊𝒏(𝒙𝟒 + 𝟓) ∗ 𝐭𝐚𝐧√𝒙 + 𝟐 + 𝒙𝟐 ∗ 𝟒𝒙 𝒄𝒐𝒔(𝒙𝟒 + 𝟓) ∗ 𝐭𝐚𝐧√𝒙 + 𝟐 𝟏 𝟏 + 𝒙𝟐 ∗ 𝒔𝒊𝒏(𝒙𝟒 + 𝟓) ∗ 𝒔𝒆𝒄𝟐 √𝒙 + 𝟐 − 𝒄𝒔𝒄𝟐 √𝒙 𝟐√𝒙 + 𝟐 𝟐√𝒙 𝟕) 𝒚 = (𝒔𝒊𝒏 𝟐𝒙 + 𝒄𝒐𝒔 𝟓𝒙 )𝟒 𝒚 = 𝟒(𝒔𝒊𝒏 𝟐𝒙 + 𝒄𝒐𝒔 𝟓𝒙 )𝟑 ∗ (𝟐 𝒄𝒐𝒔 𝟐𝒙 − 𝟓 𝒔𝒊𝒏 𝟓𝒙) 𝟖) 𝒚 = 𝒄𝒐𝒔𝟑 (𝒙𝟐 + 𝟐𝒙) + 𝒕𝒂𝒏𝟐 𝒙𝟔 + √𝒙 𝒚 = 𝟑 ∗ 𝒄𝒐𝒔𝟐 (𝒙𝟐 + 𝟐𝒙) ∗ −𝒔𝒊𝒏(𝒙𝟐 + 𝟐𝒙) ∗ (𝟐𝒙 + 𝟐) 𝟏 +𝟐 ∗ 𝒕𝒂𝒏 𝒙𝟔 + √𝒙 ∗ 𝒔𝒆𝒄𝟐 𝒙𝟔 + √𝒙 ∗ 𝟔𝒙𝟓 + 𝟐 √𝒙 𝟗) 𝒚 = 𝒔𝒆𝒄𝟒 (𝒆𝒄𝒐𝒔 𝒙 ) + 𝐥𝐧(𝐭𝐚𝐧 𝒙) + 𝐥𝐨𝐠 𝟓 (𝒙 + 𝟒)𝟐 𝒚 = 𝟒 ∗ 𝒔𝒆𝒄𝟑 (𝒆𝒄𝒐𝒔 𝒙 ) ∗ 𝐬𝐞𝐜(𝒆𝒄𝒐𝒔 𝒙 ) ∗ 𝐭𝐚𝐧(𝒆𝒄𝒐𝒔 𝒙 ) ∗ − 𝐬𝐢𝐧 𝒙 ∗ 𝒆𝒄𝒐𝒔 𝒙 𝒔𝒆𝒄𝟐 𝒙 𝟐(𝒙 + 𝟒) + + 𝐭𝐚𝐧 𝒙 𝐥𝐧 ∗ (𝒙 + 𝟒)𝟐 37  𝟑 𝟏𝟎) 𝒚 = 𝟑𝒔𝒊𝒏 𝒙 𝟓𝒙 + 𝐥𝐧(𝐬𝐞𝐜 𝒙𝟐 ) + 𝐭𝐚𝐧 𝟏 𝒙 𝟑 𝒚 = 𝟑𝒔𝒊𝒏 𝒙 𝟓𝒙 ∗ 𝐥𝐧 ∗ (𝟑𝒙𝟐 + 𝟓) 𝒄𝒐𝒔(𝒙𝟑 + 𝟓𝒙) 𝟐𝒙 ∗ 𝐬𝐞𝐜 𝒙𝟐 ∗ 𝐭𝐚𝐧 𝒙𝟐 𝟏 + + 𝐬𝐞𝐜 𝒙𝟐 𝟏 + 𝒙𝟐 𝟏𝟏) 𝒚 = 𝐬𝐢𝐧 𝟏 𝒙𝟐 + 𝐭𝐚𝐧 𝟏 √𝒙 𝟐𝒙 𝟏 𝟏 𝒚 = + ∗ √𝟏 − 𝒙𝟒 𝟏 + 𝒙 𝟐√𝒙 𝟏 𝒙𝟐 𝟏𝟐) 𝒚 = 𝒆𝐬𝐢𝐧 + 𝐥𝐧 𝒙𝟐 + 𝟑𝒙 𝟏 𝒙𝟐 𝟐𝒙 𝟐𝒙 + 𝟑 𝒚 = 𝒆𝐬𝐢𝐧 ∗ + 𝟎. 𝟓 √𝟏 − 𝒙𝟒 𝒙𝟐 + 𝟑𝒙 38