Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Logarithmic Differentiation - Type 1
Example 29 Important
Ex 5.5, 3 Important
Ex 5.5, 15
Ex 5.5, 5
Ex 5.5, 1 Important
Ex 5.5, 13
Ex 5.5, 14 Important
Ex 5.5, 16 Important
Ex 5.5, 17 Important
Ex 5.5, 18
Example 27 Important
Ex 5.5, 2
Misc 22 Important
Misc 3
Misc 7 Important
Example 41
Misc 9 Important You are here
Example 40 (i)
Logarithmic Differentiation - Type 1
Last updated at June 5, 2023 by Teachoo
Misc 9 Differentiate w.r.t. x the function, (sin𝑥−cos𝑥 )^((sin〖𝑥−cos〖𝑥)〗 〗 ), 𝜋/4 <𝑥< 3𝜋/4 Let y = (sin𝑥−cos𝑥 )^((sin〖𝑥−cos〖𝑥)〗 〗 ) Taking log on both sides log𝑦 = log (sin𝑥−cos𝑥 )^((sin〖𝑥−cos〖𝑥)〗 〗 ) log𝑦 = (sin𝑥−cos𝑥 ). 〖 log〗〖 (sin𝑥−cos𝑥 )〗 Differentiating both sides 𝑤.𝑟.𝑡.𝑥. 𝑑(log𝑦 )/𝑑𝑥 = 𝑑((sin𝑥 − cos𝑥 ). 〖 log〗(sin𝑥 − cos𝑥 ) )/𝑑𝑥 𝑑(log𝑦 )/𝑑𝑥 (𝑑𝑦/𝑑𝑦) = 𝑑((sin𝑥−cos𝑥 ). 〖 log〗〖 (sin𝑥−cos𝑥 )〗 )/𝑑𝑥 𝑑(log𝑦 )/𝑑𝑦 (𝑑𝑦/𝑑𝑥) = 𝑑((sin𝑥−cos𝑥 ). 〖 log〗〖 (sin𝑥−cos𝑥 )〗 )/𝑑𝑥 1/𝑦 . 𝑑𝑦/𝑑𝑥 = 𝑑((sin𝑥−cos𝑥 ). 〖 log〗〖 (sin𝑥−cos𝑥 )〗 " " )/𝑑𝑥 " " 1/𝑦. 𝑑𝑦/𝑑𝑥 = 𝑑(sin𝑥 − cos𝑥 )/𝑑𝑥 . 〖 log 〗(sin𝑥−cos𝑥 ) + 𝑑(〖 log〗〖 (sin𝑥 − cos𝑥 )〗 )/𝑑𝑥 .(sin𝑥−cos𝑥 ) 1/𝑦 . 𝑑𝑦/𝑑𝑥 = (cos𝑥−(−sin𝑥 )). log〖 (sin𝑥−〖 cos〗𝑥 )〗 + 1/((sin𝑥 − cos𝑥 ) ) . 𝑑(sin𝑥 − cos𝑥 )/𝑑𝑥 . (sin𝑥−〖 cos〗𝑥 ) 1/𝑦 . 𝑑𝑦/𝑑𝑥 = (cos𝑥+sin𝑥 ) . log〖 (sin𝑥−cos𝑥 )〗 + 1/((sin𝑥−cos𝑥 ) ) . (cos𝑥−(−sin𝑥 )) . (sin𝑥−cos𝑥 ) Using product rule (𝑢𝑣)’ = 𝑢’𝑣 + 𝑣’𝑢 where u = sin x − cos x & v = log (sin x − cos x) 1/𝑦 . 𝑑𝑦/𝑑𝑥 = (cos𝑥+sin𝑥 ) . log〖 (sin𝑥−cos𝑥 )〗 + 1/((sin𝑥−cos𝑥 ) ) . (cos𝑥+sin𝑥 ) . (sin𝑥−cos𝑥 ) 1/𝑦 . 𝑑𝑦/𝑑𝑥 = (cos𝑥+sin𝑥 ) . log〖 (sin𝑥−cos𝑥 )〗 + (cos𝑥+sin𝑥 ) 1/𝑦 . 𝑑𝑦/𝑑𝑥 = (cos𝑥+sin𝑥 ) . (log〖 (sin𝑥−cos𝑥 )+1〗 ) 𝑑𝑦/𝑑𝑥 = 𝑦(cos𝑥+sin𝑥 ) . (log〖 (sin𝑥−cos𝑥 )+1〗 ) 𝒅𝒚/𝒅𝒙 = (𝒔𝒊𝒏𝒙−𝒄𝒐𝒔𝒙 )^((𝒔𝒊𝒏〖𝒙−𝒄𝒐𝒔〖𝒙)〗 〗 ) (𝒄𝒐𝒔𝒙+𝒔𝒊𝒏𝒙 )(𝒍𝒐𝒈〖 (𝒔𝒊𝒏𝒙−𝒄𝒐𝒔𝒙 )+𝟏〗 )