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Misc 17 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): sin⁡〖x + cos⁡x 〗/sin⁡〖x − cos⁡x 〗 Let f (x) = sin⁡〖x + cos⁡x 〗/sin⁡〖x − cos⁡x 〗 Let u = sin x + cos x & v = sin x – cos x ∴ f(x) = 𝑢/𝑣 So, f’(x) = (𝑢/𝑣)^′ Using quotient rule f’(x) = (𝑢^′ 𝑣 −〖 𝑣〗^′ 𝑢)/𝑣^2 Finding u’ & v’ u = sin x + cos x u’ = (sin x + cos x)’ = (sin x)’ + (cos x)’ = cos x – sin x v = sin x – cos x v’= (sin x – cos x)’ = (sin x)’ – (cos x)’ = cos x – ( – sin x) = cos x + sin x Derivative of sin x = cos x Derivative of cos x = – sin x Now, f’(x) = (𝑢/𝑣)^′ = (𝑢^′ 𝑣 −〖 𝑣〗^′ 𝑢)/𝑣^2 = ( (cos⁡〖𝑥 −〖 sin〗⁡〖𝑥) (sin⁡〖𝑥 −〖 cos〗⁡〖𝑥) − (cos⁡〖𝑥 +〖 sin〗⁡〖𝑥) (sin⁡〖𝑥 +〖 cos〗⁡〖𝑥)〗 〗 〗 〗 〗 〗 〗 〗)/〖(sin⁡〖x −co𝑠 𝑥〗)〗^2 = (−(sin⁡〖𝑥 −〖 cos〗⁡〖𝑥) (sin⁡〖𝑥 −〖 cos〗⁡〖𝑥) − (sin⁡〖𝑥 + cos⁡〖𝑥) (sin⁡〖𝑥 +〖 cos〗⁡〖𝑥)〗 〗 〗 〗 〗 〗 〗 〗)/〖(sin⁡〖x − co𝑠 𝑥〗)〗^2 = (〖−(sin⁡〖x − co𝑠 𝑥〗)〗^2 − 〖(sin⁡〖x + co𝑠 𝑥〗)〗^2)/〖(sin⁡〖x − co𝑠 𝑥〗)〗^2 Using (a + b)2 = a2 + b2 + 2ab (a – b)2 = a2 + b2 – 2ab = ( − [(sin2⁡〖𝑥 +〖 cos2〗⁡〖𝑥 − 2 sin⁡〖𝑥 〖 cos〗⁡〖𝑥) + (𝑠𝑖𝑛2𝑥 + 𝑐𝑜𝑠2𝑥 + 2𝑠𝑖𝑛𝑥 cos⁡〖𝑥)]〗 〗 〗 〗 〗)/〖(sin⁡〖x − co𝑠 𝑥〗)〗^2 = ( − ( 2𝑠𝑖𝑛2𝑥 + 2𝑐𝑜𝑠2𝑥 − 0))/〖(sin⁡〖x − co𝑠 𝑥〗)〗^2 = ( −2 (𝒔𝒊𝒏𝟐𝒙 + 𝒄𝒐𝒔𝟐𝒙))/〖(sin⁡〖x − co𝑠 𝑥〗)〗^2 = ( −2 (𝟏))/〖(sin⁡〖x − co𝑠 𝑥〗)〗^2 = ( −𝟐 )/〖(𝒔𝒊𝒏⁡〖𝐱 − 𝒄𝒐𝒔 𝒙〗)〗^𝟐 (Using sin 2 x + cos 2 x = 1)

  1. Chapter 12 Class 11 Limits and Derivatives
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About the Author

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo