Example 44 - Chapter 5 Class 12 Continuity and Differentiability (Term 1)

Transcript

Example 44 Differentiate w.r.t. x, the following function: (i) √(3𝑥+2) + 1/√(2𝑥^2+ 4) Let y = √(3𝑥+2) + 1/√(2𝑥^2+ 4 ) Differentiating 𝑤.𝑟.𝑡.𝑥 𝑑𝑦/𝑑𝑥 = 𝑑(√(3𝑥 + 2) " + " 1/√(2𝑥^2 + 4 ))/𝑑𝑥 𝑑𝑦/𝑑𝑥 = 𝑑(√(3𝑥 + 2))/𝑑𝑥 + 𝑑(1/√(2𝑥^2 + 4 ))/𝑑𝑥 𝑑𝑦/𝑑𝑥 = 𝑑(√(3𝑥 + 2))/𝑑𝑥 + (𝑑(2𝑥^2 + 4)^((−1)/2))/𝑑𝑥 Calculating 𝑑(√(3𝑥 + 2))/𝑑𝑥 & (𝑑(2𝑥^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(𝟐𝒙^𝟐 + 𝟒)^((−𝟏)/𝟐))/𝒅𝒙 (𝑑(2𝑥^2 + 4)^((−1)/2))/𝑑𝑥 = (−1)/2 〖(2𝑥^2+4)〗^((−1)/( 2) −1) . 𝑑(2𝑥^2+ 4)/𝑑𝑥 = (−1)/2 (2𝑥^2+ 4)^((−3)/( 2)) . (𝑑(2𝑥^2 )/𝑑𝑥 + 𝑑(4)/𝑑𝑥) = (−1)/2 (2𝑥^2+ 4)^((−3)/( 2)) . (4𝑥+0) = (−4𝑥)/2 (2𝑥^2+ 4)^((−3)/( 2)) = (−𝟐𝒙)/(𝟐𝒙^𝟐+ 𝟒)^(𝟑/𝟐) Hence, 𝑑𝑦/𝑑𝑥 = 𝑑(√(3𝑥+2))/𝑑𝑥 + 𝑑(1/√(2𝑥^2+ 4 ))/𝑑𝑥 𝒅𝒚/𝒅𝒙 = 𝟑/(𝟐√(𝟑𝒙 + 𝟐 )) − 𝟐𝒙/(𝟐𝒙^𝟐+ 𝟒)^(𝟑/( 𝟐)) Example 44 Differentiate w.r.t. x, the following function: (ii) 𝑒^(sec^2⁡𝑥 ) + 3cos^(–1) 𝑥 Let y = 𝑒^(sec^2⁡𝑥 ) + 3cos^(–1) 𝑥 Differentiating 𝑤.𝑟.𝑡.𝑥 𝑑𝑦/𝑑𝑥 = 𝑑(𝑒^(sec^2⁡𝑥 )+ 3cos^(–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 ) "As" 𝑑(𝑒^𝑥 )/𝑑𝑥=𝑒^𝑥 & 𝑑(〖𝑐𝑜𝑠〗^(−1)⁡𝑥 )/𝑑𝑥=(−1)/√(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 〗 (As log 7 is constant) ("As (log x)′ =" 1/𝑥) ("As (log x)′ =" 1/𝑥)