Example 44 - Class 12

Transcript

Example 44 Differentiate w.r.t. x, the following function: (i) ﷮3𝑥+2﷯ + 1﷮ ﷮2 𝑥﷮2﷯+ 4﷯﷯ Let y = ﷮3𝑥+2﷯ + 1﷮ ﷮2 𝑥﷮2﷯+ 4 ﷯﷯ Differentiating 𝑤.𝑟.𝑡.𝑥 𝑑𝑦﷮𝑑𝑥﷯ = 𝑑 ﷮3𝑥+2﷯ + 1﷮ ﷮2 𝑥﷮2﷯+ 4 ﷯﷯﷯﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 𝑑 ﷮3𝑥+2﷯﷯﷮𝑑𝑥﷯ + 𝑑 1﷮ ﷮2 𝑥﷮2﷯+ 4 ﷯﷯﷯﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 𝑑 ﷮3𝑥 + 2﷯﷯﷮𝑑𝑥﷯ + 𝑑 2 𝑥﷮2﷯ + 4﷯﷮ −1﷮2﷯﷯﷮𝑑𝑥﷯ Calculating 𝑑 ﷮3𝑥 + 2﷯﷯﷮𝑑𝑥﷯ & 𝑑 2 𝑥﷮2﷯ + 4﷯﷮ −1﷮2﷯﷯﷮𝑑𝑥﷯ separately Calculating 𝐝 ﷮𝟑𝐱 + 𝟐﷯﷯﷮𝒅𝒙﷯ 𝑑 ﷮3𝑥 + 2﷯﷯﷮𝑑𝑥﷯ = 1﷮2 ﷮3𝑥 + 2 ﷯﷯ × 𝑑 3𝑥 + 2﷯﷮𝑑𝑥﷯ = 1﷮2 ﷮3𝑥 + 2 ﷯﷯ × 3+0﷯ = 3﷮2 ﷮3𝑥 + 2 ﷯﷯ Calculating 𝒅 𝟐 𝒙﷮𝟐﷯ + 𝟒﷯﷮ −𝟏﷮𝟐﷯﷯﷮𝒅𝒙﷯ 𝑑 2𝑥 + 4﷯﷮ −1﷮2﷯﷯﷮𝑑𝑥﷯ = −1﷮2﷯ (2𝑥+4)﷮ −1﷮ 2﷯ −1﷯ . 𝑑 2 𝑥﷮2﷯+ 4﷯﷮𝑑𝑥﷯ = −1﷮2﷯ 2 𝑥﷮2﷯+ 4﷯﷮ −3﷮ 2﷯﷯ . 𝑑 2 𝑥﷮2﷯+ 4﷯﷮𝑑𝑥﷯ = −1﷮2﷯ 2 𝑥﷮2﷯+ 4﷯﷮ −3﷮ 2﷯﷯ . 𝑑 2 𝑥﷮2﷯﷯﷮𝑑𝑥﷯ + 𝑑 4﷯﷮𝑑𝑥﷯﷯ = −1﷮2﷯ 2 𝑥﷮2﷯+ 4﷯﷮ −3﷮ 2﷯﷯ . 4𝑥+0﷯ = −4𝑥﷮2﷯ 2 𝑥﷮2﷯+ 4﷯﷮ −3﷮ 2﷯﷯ = −2𝑥﷮ 2 𝑥﷮2﷯+ 4﷯﷮ 3﷮2﷯﷯﷯ Hence, 𝑑𝑦﷮𝑑𝑥﷯ = 𝑑 ﷮3𝑥+2﷯﷯﷮𝑑𝑥﷯ + 𝑑 1﷮ ﷮2 𝑥﷮2﷯+ 4 ﷯﷯﷯﷮𝑑𝑥﷯ 𝒅𝒚﷮𝒅𝒙﷯ = 𝟑﷮𝟐 ﷮𝟑𝒙 + 𝟐 ﷯﷯ − 𝟐𝒙﷮ 𝟐 𝒙﷮𝟐﷯+ 𝟒﷯﷮ −𝟑﷮ 𝟐﷯﷯﷯ Example 44 Differentiate w.r.t. x, the following function: (ii) 𝑒﷮ sec﷮2﷯﷮𝑥﷯ ﷯ + 3 cos﷮–1﷯ 𝑥 Let y = 𝑒﷮ sec﷮2﷯﷮𝑥﷯ ﷯ + 3 cos﷮–1﷯ 𝑥 Differentiating 𝑤.𝑟.𝑡.𝑥 𝑑𝑦﷮𝑑𝑥﷯ = 𝑑 𝑒﷮ sec﷮2﷯﷮𝑥﷯ ﷯+ 3 cos﷮–1﷯ 𝑥 ﷯﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 𝑑 𝑒﷮ sec﷮2﷯﷮𝑥﷯ ﷯﷯﷮𝑑𝑥﷯ + 𝒅 𝟑 𝒄𝒐𝒔﷮–𝟏﷯ 𝒙﷯﷮𝒅𝒙﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 𝑒﷮ sec﷮2﷯﷮𝑥﷯ ﷯ 𝑑 sec﷮2﷯﷮𝑥﷯﷯﷮𝑑𝑥﷯ + 3. −𝟏﷮ ﷮𝟏 − 𝒙﷮𝟐﷯﷯﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 𝑒﷮ sec﷮2﷯﷮𝑥﷯ ﷯. 2 sec 𝑥 . 𝒅 𝒔𝒆𝒄 ﷮𝒙﷯﷯﷮𝒅𝒙﷯ − 3﷮ ﷮1 − 𝑥﷮2﷯﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 𝑒﷮ sec﷮2﷯﷮𝑥﷯ ﷯. 2 sec 𝑥 . 𝒔𝒆𝒄﷮𝒙﷯ . 𝒕𝒂𝒏﷮𝒙﷯ − 3﷮ ﷮1 − 𝑥﷮2﷯﷯﷯ 𝒅𝒚﷮𝒅𝒙﷯ = 𝟐 𝒆﷮ 𝒔𝒆𝒄﷮𝟐﷯﷮𝒙﷯ ﷯. 𝒔𝒆𝒄﷮𝟐﷯𝒙 . 𝒕𝒂𝒏﷮𝒙﷯ − 𝟑﷮ ﷮𝟏 − 𝒙﷮𝟐﷯﷯﷯ Example 44 Differentiate w.r.t. x, the following function: (iii) log7 (log x) y = log7 (log x) But we do not solve base 7. So, changing base y = log7 ( log x) y = 𝐥𝐨𝐠﷮( 𝐥𝐨𝐠﷮𝒙)﷯﷯﷮ 𝐥𝐨𝐠﷮𝟕﷯﷯ Now, differentiating 𝑑𝑦﷮𝑑𝑥﷯= 𝑑 ( log﷮( log﷮7)﷯﷯﷮log 7﷯﷯﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯= 𝟏﷮ 𝐥𝐨𝐠﷮𝟕﷯﷯ 𝑑 log﷮( log﷮𝑥﷯﷯﷯) ﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯= 1﷮ log﷮7﷯﷯. 1﷮ log﷮𝑥﷯﷯ 𝑑( log﷮𝑥﷯)﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯= 1﷮ log﷮7﷯﷯. 1﷮ log﷮𝑥﷯﷯. 1﷮𝑥﷯ 𝒅𝒚﷮𝒅𝒙﷯= 𝟏﷮ 𝐱 𝒍𝒐𝒈﷮𝟕 𝐥𝐨𝐠﷮𝒙﷯﷯﷯