Check sibling questions

Example 15 - Find point at which tangent to y = root 4x-3 - 1

Example 15 - Chapter 6 Class 12 Application of Derivatives - Part 2
Example 15 - Chapter 6 Class 12 Application of Derivatives - Part 3


Transcript

Example 15 Find the point at which the tangent to the curve 𝑦 = √(4π‘₯βˆ’3)βˆ’1 has its slope 2/3 . Given, Slope of the tangent to the curve is 2/3 We know that Slope of tangent = π’…π’š/𝒅𝒙 2/3 = 𝑑𝑦/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = 2/3 𝑑(√(4π‘₯ βˆ’ 3) βˆ’ 1)/𝑑π‘₯ = 2/3 1/(2√(4π‘₯ βˆ’ 3)) Γ—4βˆ’0 = 2/3 2/√(4π‘₯ βˆ’ 3) = 2/3 3 = √(4π‘₯βˆ’3) √(πŸ’π’™βˆ’πŸ‘) = 3 Squaring both sides 4x βˆ’ 3 = 9 4x = 12 x = 3 Finding y for x = 3 𝑦=√(4π‘₯βˆ’3) βˆ’ 1 =√(12βˆ’3)βˆ’1 =√9βˆ’1 =3βˆ’1 =𝟐 Hence, the required point is (πŸ‘, 𝟐)

Davneet Singh's photo - Teacher, Engineer, Marketer

Made by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.