Question 2 - Examples - Chapter 6 Class 12 Application of Derivatives

Last updated at August 17, 2026 by

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

Question 2 Find the point at which the tangent to the curve 𝑦 = √(4𝑥−3)−1 has its slope 2/3 . Given, Slope of the tangent to the curve is 2/3 We know that Slope of tangent = 𝒅𝒚/𝒅𝒙 2/3 = 𝑑𝑦/𝑑𝑥 𝑑𝑦/𝑑𝑥 = 2/3 𝑑(√(4𝑥 − 3) − 1)/𝑑𝑥 = 2/3 1/(2√(4𝑥 − 3)) ×4−0 = 2/3 2/√(4𝑥 − 3) = 2/3 3 = √(4𝑥−3) √(𝟒𝒙−𝟑) = 3 Squaring both sides 4x − 3 = 9 4x = 12 x = 3 Finding y for x = 3 𝑦=√(4𝑥−3) − 1 =√(12−3)−1 =√9−1 =3−1 =𝟐 Hence, the required point is (𝟑, 𝟐)

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