Ex 5.5, 9 - Class 12

Transcript

Ex 5.5, 9 (Method 1) Differentiate the functions in, 𝑥﷮ sin﷮𝑥﷯﷯+ ( sin﷮𝑥﷯)﷮ cos﷮𝑥﷯﷯ Let y = 𝑥﷮ sin﷮𝑥﷯﷯+ ( sin﷮𝑥﷯)﷮ cos﷮𝑥 ﷯﷯ Let 𝑢 = 𝑥﷮ sin﷮𝑥﷯﷯ & 𝑣 = ( sin﷮𝑥﷯)﷮ cos﷮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﷮ sin﷮𝑥﷯﷯ Taking log both sides log﷮𝑢﷯= log﷮ 𝑥﷮ sin﷮𝑥﷯﷯﷯ log﷮𝑢﷯= sin﷮𝑥﷯. log﷮ 𝑥﷯ Differentiating both sides 𝑤.𝑟.𝑡.𝑥. 𝑑( log﷮𝑢﷯)﷮𝑑𝑥﷯ = 𝑑 sin﷮𝑥﷯. log﷮ 𝑥﷯﷯﷮𝑑𝑥﷯ 𝑑( log﷮𝑢﷯)﷮𝑑𝑥﷯ . 𝑑𝑢﷮𝑑𝑢﷯ = 𝑑 sin﷮𝑥﷯. log﷮ 𝑥﷯﷯﷮𝑑𝑥﷯ 1﷮𝑢﷯ . 𝑑𝑢﷮𝑑𝑥﷯ = 𝑑 sin﷮𝑥﷯ . log﷮ 𝑥﷯﷯﷮𝑑𝑥﷯ 1﷮𝑢﷯ . 𝑑𝑢﷮𝑑𝑥﷯ = 𝑑 sin﷮𝑥﷯﷯﷮𝑑𝑥﷯ . log⁡𝑥 + 𝑑 log﷮𝑥﷯﷯﷮𝑑𝑥﷯ . sin 𝑥 1﷮𝑢﷯ . 𝑑𝑢﷮𝑑𝑥﷯ = cos﷮𝑥﷯ . log⁡𝑥 + 1﷮𝑥﷯ . sin 𝑥 1﷮𝑢﷯ . 𝑑𝑢﷮𝑑𝑥﷯ = cos﷮𝑥﷯ . log⁡𝑥 + sin 𝑥 ﷮𝑥﷯ 𝑑𝑢﷮𝑑𝑥﷯ = u cos﷮𝑥 . log﷮𝑥+ sin﷮𝑥﷯﷮𝑥﷯﷯﷯﷯ 𝑑𝑢﷮𝑑𝑥﷯ = 𝑥﷮ sin﷮𝑥﷯﷯ cos﷮𝑥 . log﷮𝑥+ sin﷮𝑥﷯﷮𝑥﷯﷯﷯﷯. 𝑑𝑢﷮𝑑𝑥﷯ = 𝑥﷮ sin﷮𝑥﷯﷯ sin﷮𝑥﷯﷮𝑥﷯ +cos﷮𝑥 . log﷮𝑥﷯﷯﷯ Calculating 𝒅𝒗﷮𝒅𝒙﷯ 𝑣= sin﷮𝑥﷯﷯﷮ cos﷮𝑥﷯﷯ Taking log both sides log⁡𝑣 = log⁡ . sin﷮𝑥﷯﷮cos 𝑥﷯ log﷮𝑣﷯= cos﷮𝑥﷯. log sin﷮𝑥﷯ Differentiating both sides 𝑤.𝑟.𝑡.𝑥. 𝑑( log﷮𝑣﷯)﷮𝑑𝑥﷯ = 𝑑 cos﷮𝑥﷯. log sin﷮𝑥﷯﷯﷮𝑑𝑥﷯ 𝑑( log﷮𝑣﷯)﷮𝑑𝑥﷯ . 𝑑𝑣﷮𝑑𝑣﷯ = 𝑑 cos﷮𝑥﷯. log sin﷮𝑥﷯﷯﷮𝑑𝑥﷯ 𝑑( log﷮𝑣﷯)﷮𝑑𝑣﷯ . 𝑑𝑣﷮𝑑𝑥﷯ = 𝑑 cos﷮𝑥﷯. log sin﷮𝑥﷯﷯﷮𝑑𝑥﷯ 1﷮𝑣﷯ . 𝑑𝑣﷮𝑑𝑥﷯ = 𝑑 cos﷮𝑥﷯. log sin﷮𝑥﷯﷯﷮𝑑𝑥﷯ 1﷮𝑣﷯ . 𝑑𝑣﷮𝑑𝑥﷯ = 𝑑 cos﷮𝑥﷯﷯﷮𝑑𝑥﷯ . log⁡sin⁡𝑥 + 𝑑 log﷮ sin﷮𝑥﷯﷯﷯﷮𝑑𝑥﷯ . cos⁡𝑥 1﷮𝑣﷯ . 𝑑𝑣﷮𝑑𝑥﷯ = − sin﷮𝑥﷯ . log⁡sin⁡𝑥 + 1﷮ sin﷮𝑥﷯﷯ . 𝑑 sin﷮𝑥﷯﷯﷮𝑑𝑥﷯﷯ . cos⁡𝑥 1﷮𝑣﷯ . 𝑑𝑣﷮𝑑𝑥﷯ = − sin﷮𝑥﷯ . log⁡sin⁡𝑥 + 1﷮ sin﷮𝑥﷯﷯ . cos﷮𝑥﷯﷯ . cos⁡𝑥 1﷮𝑣﷯ . 𝑑𝑣﷮𝑑𝑥﷯ = − sin﷮𝑥﷯ . log⁡sin⁡𝑥 + cot﷮𝑥﷯﷯ . cos⁡𝑥 𝑑𝑣﷮𝑑𝑥﷯ = 𝑣 − sin﷮𝑥﷯ . log⁡sin⁡𝑥 + cot﷮𝑥﷯. cos⁡𝑥﷯ 𝑑𝑣﷮𝑑𝑥﷯ = sin﷮𝑥﷯﷯﷮ cos﷮𝑥﷯﷯ co𝑠﷮𝑥﷯. co𝑡⁡𝑥− sin﷮𝑥﷯ . log⁡sin⁡𝑥﷯ Now 𝑑𝑦﷮𝑑𝑥﷯ = 𝑑𝑢﷮𝑑𝑥﷯ + 𝑑𝑣﷮𝑑𝑥﷯ Putting value of 𝑑𝑢﷮𝑑𝑥﷯ & 𝑑𝑣﷮𝑑𝑥﷯ 𝒅𝒚﷮𝒅𝒙﷯ = 𝒙﷮ 𝐬𝐢𝐧﷮𝒙﷯﷯ 𝐬𝐢𝐧﷮𝒙﷯﷮𝒙﷯+ 𝐜𝐨𝐬﷮𝒙 . 𝐥𝐨𝐠﷮𝒙﷯﷯﷯ + 𝐬𝐢𝐧﷮𝒙﷯﷯﷮ 𝐜𝐨𝐬﷮𝒙﷯﷯ 𝒄𝒐𝒔﷮𝒙﷯. 𝒄𝒐𝒕⁡𝒙− 𝒔𝒊𝒏﷮𝒙﷯ . 𝒍𝒐𝒈⁡𝒔𝒊𝒏⁡﷯