Question 2 - Tangents and Normals (using Differentiation) - Chapter 6 Class 12 Application of Derivatives

Last updated at April 16, 2024 by Teachoo

Ex 6.3, 2 - Find slope of tangent y = x-1/x-2, at x = 10

Ex 6.3,2 - Chapter 6 Class 12 Application of Derivatives - Part 2

Transcript

Question 2 Find the slope of the tangent to the curve 𝑦=(𝑥 − 1)/(𝑥 − 2) , 𝑥≠2 at 𝑥=10 𝑦=(𝑥 − 1)/(𝑥 − 2) 𝑥≠2 We know that Slope of tangent is 𝑑𝑦/𝑑𝑥 𝑑𝑦/𝑑𝑥= 𝑑((𝑥 − 1)/(𝑥 − 2))" " /𝑑𝑥 𝑑𝑦/𝑑𝑥=((𝑥 − 1)^′ (𝑥 − 2) − (𝑥 − 2)^′ (𝑥 − 1))/(𝑥 − 2)^2 Using Quotient Rule As (𝑢/𝑣)^′=(𝑢^′ 𝑣 −〖 𝑣〗^′ 𝑢)/𝑣^2 𝑑𝑦/𝑑𝑥=(1 (𝑥 − 2) −1 (𝑥 − 1))/(𝑥 − 2)^2 𝑑𝑦/𝑑𝑥=(𝑥 − 2 − 𝑥 + 1)/(𝑥 − 2)^2 So, 𝑑𝑦/𝑑𝑥=(− 1)/(𝑥 − 2)^2 Putting 𝑥=10 (𝑑𝑦/𝑑𝑥)_(𝑥 = 10)=(− 1)/(10 − 2)^2 =(− 1)/〖 8〗^2 =(− 1)/64 Hence Slope of a tangent at 𝑥 = 10 is (−𝟏)/𝟔𝟒

Next: Question 3 Important → Go Ad-free

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.

Class 6

Class 7

Class 8

Class 9

Class 10

Class 11

Class 12

Courses

Interesting

Popular