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Last updated at Nov. 30, 2019 by Teachoo

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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 + cosx 〗/sin〖x − cosx 〗 Let f (x) = sin〖x + cosx 〗/sin〖x − cosx 〗 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)

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.