Limit of (1-cos(x))/x as x approaches 0 | Derivative rules | AP Calculus AB | Khan Academy
Limit of (1-cos(x))/x as x approaches 0 | Derivative rules | AP Calculus AB | Khan Academy

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Question 1 Deleted for CBSE Board 2024 Exams

Question 2 Important Deleted for CBSE Board 2024 Exams

Question 3 Deleted for CBSE Board 2024 Exams

Question 4 Important Deleted for CBSE Board 2024 Exams

Question 5 Deleted for CBSE Board 2024 Exams

Question 6 Important Deleted for CBSE Board 2024 Exams

Last updated at June 2, 2023 by Teachoo

Example 40 (Method 1) Differentiate the following w.r.t. x. (ii) tan −1 (sin𝑥/( 1 +〖 cos〗〖𝑥 〗 )) Let 𝑓(𝑥) = tan −1 (𝒔𝒊𝒏𝒙/( 1 +〖 𝒄𝒐𝒔〗〖𝒙 〗 )) 𝑓(𝑥) = tan −1 ((𝟐 〖𝐬𝐢𝐧 〗〖𝒙/𝟐〗 〖 𝐜𝐨𝐬 〗〖𝒙/𝟐〗)/( 1+ (𝟐 〖𝐜𝐨𝐬〗^𝟐〖 𝒙/𝟐〗 − 𝟏))) = tan −1 ((2 〖sin 〗〖𝑥/2〗 〖 cos 〗〖𝑥/2 〗)/( 2 cos^2〖 𝑥/2〗 )) = tan −1 ((2 〖sin 〗〖𝑥/2〗)/( 2 cos〖𝑥/2〗 )) We know that sin 2θ = 2 sin θ cos θ Replacing θ by 𝜃/2 sin θ = 2 𝒔𝒊𝒏〖𝜽/𝟐〗 𝒄𝒐𝒔〖𝜽/𝟐〗 and cos 2θ = 2cos2 θ – 1 Replacing θ by 𝜃/2 cos θ = 2cos2 𝜽/𝟐 – 1 = tan −1 (〖sin 〗〖𝑥/2〗/( cos〖 𝑥/2〗 )) = tan −1 (〖tan 〗〖𝑥/2〗 ) = 𝒙/𝟐 𝒇(𝒙) = 𝒙/𝟐 Differentiating 𝑤.𝑟.𝑡.𝑥 𝑓’(𝑥) = 1/2 (𝑑(𝑥))/𝑑𝑥 𝒇’(𝒙) = 𝟏/𝟐 (As tan^(−1)〖(tan𝜃)〗 =𝜃) Example 40 (Method 2) Differentiate the following w.r.t. x. (ii) tan −1 (sin𝑥/( 1 +〖 cos〗〖𝑥 〗 )) Let 𝑓(𝑥) = tan −1 (𝒔𝒊𝒏𝒙/( 1 +〖 𝒄𝒐𝒔〗〖𝒙 〗 )) Differentiating w.r.t x 𝑓^′ (𝑥) = 1/(1 + (𝒔𝒊𝒏𝒙/( 1 +〖 𝒄𝒐𝒔〗〖𝒙 〗 ))^2 ) (𝒔𝒊𝒏𝒙/( 1 +〖 𝒄𝒐𝒔〗〖𝒙 〗 ))^′ = 1/(((1 + cos𝑥 )^2 + sin^2𝑥)/(1 + cos𝑥 )^2 ) (((𝒔𝒊𝒏𝒙 )^′ (𝟏 + 𝒄𝒐𝒔 𝒙)−(𝟏 + 𝒄𝒐𝒔 𝒙)^′ 𝒔𝒊𝒏 𝒙)/( (1 +〖 𝒄𝒐𝒔〗𝒙 )^𝟐 )) = (1 + cos𝑥 )^2/((1 + cos𝑥 )^2 + sin^2𝑥 ) ((𝒄𝒐𝒔 𝒙(𝟏 + 𝒄𝒐𝒔 𝒙). − (− 𝒔𝒊𝒏 𝒙)𝒔𝒊𝒏 𝒙)/( (1 +〖 𝒄𝒐𝒔〗𝒙 )^𝟐 ))= (𝑐𝑜𝑠 𝑥 + 𝒄𝒐𝒔^𝟐 𝒙 + 𝒔𝒊𝒏^𝟐 𝒙)/(1 + 〖𝐜𝐨𝐬〗^𝟐𝒙 + 2 cos𝑥 + 〖𝒔𝒊𝒏〗^𝟐𝒙 ) = (𝒄𝒐𝒔 𝒙 +𝟏)/(1 + 1 + 2 cos𝑥 ) = (𝒄𝒐𝒔 𝒙 + 𝟏)/(2 + 2 cos𝑥 ) = (1 + 𝒄𝒐𝒔 𝒙)/(2(1 + cos𝑥) ) = 1/2

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