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

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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)

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.