Slide24.JPG

Slide25.JPG
Slide26.JPG
Slide27.JPG

 

 

Go Ad-free

Transcript

Example 39 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))

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.