Ex 5.5, 11 - Class 12

Transcript

Ex 5.5, 11 Differentiate the functions in, 𝑥 𝑐𝑜𝑠﷮𝑥﷯﷯ ﷮𝑥﷯ + 𝑥 𝑠𝑖𝑛﷮𝑥﷯﷯ ﷮ 1﷮𝑥﷯﷯ 𝑦 = 𝑥 𝑐𝑜𝑠﷮𝑥﷯﷯ ﷮𝑥﷯ + 𝑥 𝑠𝑖𝑛﷮𝑥﷯﷯ ﷮ 1﷮𝑥﷯﷯ Let 𝑢 = 𝑥 𝑐𝑜𝑠﷮𝑥﷯﷯ ﷮𝑥﷯ , 𝑣 = 𝑥 𝑠𝑖𝑛﷮𝑥﷯﷯ ﷮ 1﷮𝑥﷯﷯ 𝑦 = 𝑢+𝑣 Differentiating both sides 𝑤.𝑟.𝑡.𝑥. 𝑑𝑦﷮𝑑𝑥﷯ = 𝑑 𝑢 + 𝑣﷯﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 𝑑𝑢﷮𝑑𝑥﷯ + 𝑑𝑣﷮𝑑𝑥﷯ Calculating 𝒅𝒖﷮𝒅𝒙﷯ 𝑢 = 𝑥 𝑐𝑜𝑠﷮𝑥﷯﷯ ﷮𝑥﷯ Taking log both sides . log⁡𝑢 = log 𝑥 𝑐𝑜𝑠﷮𝑥﷯﷯ ﷮𝑥﷯ log⁡𝑢 = 𝑥 . log 𝑥 𝑐𝑜𝑠﷮𝑥﷯﷯ ﷮𝑥﷯ log⁡𝑢 = 𝑥 log 𝑥+ log﷮ cos﷮𝑥﷯﷯﷯ log⁡𝑢 = 𝑥 . log 𝑥 + 𝑥 . log﷮ cos﷮𝑥﷯﷯ Differentiating both sides 𝑤.𝑟.𝑡.𝑥. 𝑑 log﷮𝑢﷯﷯﷮𝑑𝑥﷯ = 𝑑 𝑥 . log 𝑥 + 𝑥 . log﷮ cos﷮𝑥﷯﷯﷯﷮𝑑𝑥﷯ 𝑑 log﷮𝑢﷯﷯﷮𝑑𝑥﷯ . 𝑑𝑢﷮𝑑𝑢﷯ = 𝑑 𝑥 . log 𝑥 + 𝑥 . log﷮ cos﷮𝑥﷯﷯﷯﷮𝑑𝑥﷯ 1﷮𝑢﷯ . 𝑑𝑢﷮𝑑𝑥﷯ = 𝑑 𝑥 . log 𝑥 + 𝑥 . log﷮ cos﷮𝑥﷯﷯﷯﷮𝑑𝑥﷯ 1﷮𝑢﷯ . 𝑑𝑢﷮𝑑𝑥﷯ = 𝑑 𝑥 . log 𝑥﷯﷮𝑑𝑥﷯ + 𝑑 𝑥 . log﷮ cos﷮𝑥﷯﷯﷯﷮𝑑𝑥﷯ 1﷮𝑢﷯ 𝑑𝑢﷮𝑑𝑥﷯﷯= 𝑑 𝑥﷯﷮𝑑𝑥﷯ .log 𝑥+ 𝑑 log 𝑥﷯﷮𝑑𝑥﷯. 𝑥﷯ + 𝑑 𝑥﷯﷮𝑑𝑥﷯ .log cos﷮𝑥﷯+ 𝑑 log cos﷮𝑥﷯﷯﷮𝑑𝑥﷯. 𝑥﷯ 1﷮𝑢﷯ 𝑑𝑢﷮𝑑𝑥﷯﷯ = 1 .log 𝑥+ 1﷮𝑥﷯. 𝑥﷯ + 1.log cos﷮𝑥﷯+ 1﷮ cos﷮𝑥﷯﷯. 𝑑 cos﷮𝑥﷯﷯﷮𝑑𝑥﷯. 𝑥﷯ 1﷮𝑢﷯ 𝑑𝑢﷮𝑑𝑥﷯﷯ = log 𝑥+1﷯ + log cos﷮𝑥﷯+ 1﷮ cos﷮𝑥﷯﷯. − sin﷮𝑥﷯﷯. 𝑥﷯ 1﷮𝑢﷯ 𝑑𝑢﷮𝑑𝑥﷯﷯ = log 𝑥+1﷯ + log cos﷮𝑥﷯− sin﷮𝑥﷯﷮ cos﷮𝑥﷯﷯ . 𝑥﷯ 1﷮𝑢﷯ 𝑑𝑢﷮𝑑𝑥﷯﷯ = log 𝑥+1 + log cos﷮𝑥﷯−tan 𝑥 . 𝑥 1﷮𝑢﷯ 𝑑𝑢﷮𝑑𝑥﷯﷯ = 1. tan 𝑥+ log 𝑥+log cos﷮𝑥﷯ 𝑑𝑢﷮𝑑𝑥﷯ = u 1−𝑥 tan﷮𝑥﷯+ log﷮𝑥+ log﷮ cos﷮𝑥﷯﷯﷯﷯ Putting value of 𝑢 𝑑𝑢﷮𝑑𝑥﷯ = 𝑥 cos﷮𝑥﷯﷯﷮𝑥﷯ 1−𝑥 tan﷮𝑥﷯+ 𝒍𝒐𝒈﷮𝒙﷯+ 𝒍𝒐𝒈﷮ 𝒄𝒐𝒔﷮𝒙﷯﷯﷯ 𝑑𝑢﷮𝑑𝑥﷯ = 𝑥 cos﷮𝑥﷯﷯﷮𝑥﷯ 1−𝑥 tan﷮𝑥﷯+ 𝒍𝒐𝒈﷮ 𝒙 𝒄𝒐𝒔﷮𝒙﷯﷯﷯﷯ Calculating 𝒅𝒗﷮𝒅𝒙﷯ 𝑣= 𝑥 𝑠𝑖𝑛﷮𝑥﷯﷯ ﷮ 1﷮𝑥﷯﷯ Taking log both sides log﷮𝑣﷯= log﷮ 𝑥 𝑠𝑖𝑛﷮𝑥﷯﷯ ﷮ 1﷮𝑥﷯﷯﷯ log﷮𝑣﷯= 1﷮𝑥﷯ log 𝑥 𝑠𝑖𝑛﷮𝑥﷯﷯ log﷮𝑣﷯= 1﷮𝑥﷯ log 𝑥+ log ﷮ 𝑠𝑖𝑛﷮𝑥﷯﷯﷯ log﷮𝑣﷯= 1﷮𝑥﷯ . log 𝑥 + 1﷮𝑥﷯ log ﷮ 𝑠𝑖𝑛﷮𝑥﷯﷯ Differentiating both sides 𝑤.𝑟.𝑡.𝑥. 𝑑( log﷮𝑣﷯)﷮𝑑𝑥﷯ = 𝑑 1﷮𝑥﷯ . log 𝑥 + 1﷮𝑥﷯ log ﷮ 𝑠𝑖𝑛﷮𝑥﷯﷯﷯﷮𝑑𝑥﷯ 𝑑( log﷮𝑣﷯)﷮𝑑𝑥﷯ . 𝑑𝑣﷮𝑑𝑣﷯ = 𝑑 1﷮𝑥﷯ . log 𝑥﷯﷮𝑑𝑥﷯ + 𝑑 1﷮𝑥﷯ log ﷮ 𝑠𝑖𝑛﷮𝑥﷯﷯﷯﷮𝑑𝑥﷯ 𝑑( log﷮𝑣﷯)﷮𝑑𝑣﷯ . 𝑑𝑣﷮𝑑𝑥﷯ = 𝑑﷮𝑑𝑥﷯ 1﷮𝑥﷯ . log﷮𝑥﷯﷯ + 𝑑﷮𝑑𝑥﷯ 1﷮𝑥﷯ . log﷮ sin﷮𝑥﷯﷯﷯ 1﷮𝑣﷯ × 𝑑𝑣﷮𝑑𝑥﷯ = 𝑑﷮𝑑𝑥﷯ 1﷮𝑥﷯ . log﷮𝑥﷯﷯ + 𝑑﷮𝑑𝑥﷯ 1﷮𝑥﷯ . log﷮ sin﷮𝑥﷯﷯﷯ 1﷮𝑣﷯ × 𝑑𝑣﷮𝑑𝑥﷯ = 𝑑﷮𝑑𝑥﷯ log﷮𝑥﷯﷮𝑥﷯﷯ + 𝑑﷮𝑑𝑥﷯ log﷮ sin﷮𝑥﷯﷯﷮𝑥﷯﷯ 1﷮𝑣﷯ × 𝑑𝑣﷮𝑑𝑥﷯ = 𝑑 log﷮𝑥﷯﷯﷮𝑑𝑥﷯ . 𝑥 − 𝑑 𝑥﷯﷮𝑑𝑥﷯ . log﷮𝑥﷯﷮ 𝑥﷮2﷯﷯﷯ + 𝑑 log sin﷮𝑥﷯﷯﷮𝑑𝑥﷯ . 𝑥 − 𝑑 𝑥﷯﷮𝑑𝑥﷯ . log 𝑠𝑖𝑛﷮𝑥﷯﷮ 𝑥﷮2﷯﷯﷯ 1﷮𝑣﷯ × 𝑑𝑣﷮𝑑𝑥﷯ = 1﷮𝑥﷯ . 𝑥 − 1 . log﷮𝑥﷯﷮ 𝑥﷮2﷯﷯﷯ + 1﷮ sin﷮𝑥﷯﷯ . 𝑑 sin﷮𝑥﷯﷯﷮𝑑𝑥﷯ . 𝑥 − log 𝑠𝑖𝑛﷮𝑥﷯﷮ 𝑥﷮2﷯﷯﷯ 1﷮𝑣﷯ × 𝑑𝑣﷮𝑑𝑥﷯ = 1 − log﷮𝑥﷯﷮ 𝑥﷮2﷯﷯﷯ + 1﷮ sin﷮𝑥﷯﷯ . cos﷮𝑥﷯ . 𝑥 − log 𝑠𝑖𝑛﷮𝑥﷯﷮ 𝑥﷮2﷯﷯﷯ 1﷮𝑣﷯ × 𝑑𝑣﷮𝑑𝑥﷯ = 1 − log﷮𝑥﷯﷮ 𝑥﷮2﷯﷯﷯ + cot﷮𝑥 . 𝑥 − log﷮ sin﷮𝑥﷯﷯﷯﷮ 𝑥﷮2﷯﷯ 1﷮𝑣﷯ × 𝑑𝑣﷮𝑑𝑥﷯ = 1 − log﷮𝑥﷯ + cot﷮𝑥 . 𝑥 − log﷮ sin﷮𝑥﷯﷯﷯﷮ 𝑥﷮2﷯﷯﷯ 1﷮𝑣﷯ × 𝑑𝑣﷮𝑑𝑥﷯ = 𝑥 cot﷮𝑥 +1 − log﷮ 𝑥﷮− log﷮ sin﷮𝑥﷯﷯﷯﷯﷯﷮ 𝑥﷮2﷯﷯﷯ 1﷮𝑣﷯ × 𝑑𝑣﷮𝑑𝑥﷯ = 𝑥 cot﷮𝑥 +1 − log﷮ 𝑥﷮+ log﷮ sin﷮𝑥﷯﷯﷯﷯﷯﷯﷮ 𝑥﷮2﷯﷯﷯ 1﷮𝑣﷯ × 𝑑𝑣﷮𝑑𝑥﷯ = 𝑥 cot 𝑥 + 1 − 𝒍𝒐𝒈﷮ 𝒙 𝒔𝒊𝒏﷮𝒙﷯﷯﷯﷮ 𝑥﷮2﷯﷯ 1﷮𝑣﷯ × 𝑑𝑣﷮𝑑𝑥﷯ = 𝑥 cot 𝑥 + 1 − 𝒍𝒐𝒈﷮ 𝒙 𝒔𝒊𝒏﷮𝒙﷯﷯﷯﷮ 𝑥﷮2﷯﷯ 𝑑𝑣﷮𝑑𝑥﷯ = 𝑣 𝑥 cot﷮𝑥 +1 − log﷮𝑥 sin﷮𝑥﷯﷯﷯﷮ 𝑥﷮2﷯﷯﷯ 𝑑𝑣﷮𝑑𝑥﷯ = 𝑥 sin﷮𝑥﷯﷯﷮ 1﷮𝑥﷯﷯ 𝑥 cot﷮𝑥 +1 − log 𝑥 sin﷮𝑥﷯﷯﷯﷮ 𝑥﷮2﷯﷯﷯ Now, Hence 𝑑𝑦﷮𝑑𝑥﷯ = 𝑑𝑢﷮𝑑𝑥﷯ + 𝑑𝑣﷮𝑑𝑥﷯ Putting value of 𝑑𝑢﷮𝑑𝑥﷯ & 𝑑𝑣﷮𝑑𝑥﷯ 𝒅𝒚﷮𝒅𝒙﷯ = 𝒙 𝒔𝒊𝒏﷮𝒙﷯﷯﷮𝒙﷯ 𝟏 − 𝒙 𝒕𝒂𝒏﷮𝒙﷯+ 𝒍𝒐𝒈﷮ 𝒙 𝒄𝒐𝒔﷮𝒙﷯﷯﷯﷯ + 𝒙 𝒔𝒊𝒏﷮𝒙﷯﷯﷮ 𝟏﷮𝒙﷯﷯ 𝒙 𝒄𝒐𝒕﷮𝒙 +𝟏 − 𝐥𝐨𝐠 𝒙 𝒔𝒊𝒏﷮𝒙﷯﷯﷯﷮ 𝒙﷮𝟐﷯﷯﷯