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

Last updated at December 16, 2024 by

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

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Question 2 Find the slope of the tangent to the curve š‘¦=(š‘„ āˆ’ 1)/(š‘„ āˆ’ 2) , š‘„ā‰ 2 at š‘„=10 š‘¦=(š‘„ āˆ’ 1)/(š‘„ āˆ’ 2) š‘„ā‰ 2 We know that Slope of tangent is š‘‘š‘¦/š‘‘š‘„ š‘‘š‘¦/š‘‘š‘„= š‘‘((š‘„ āˆ’ 1)/(š‘„ āˆ’ 2))" " /š‘‘š‘„ š‘‘š‘¦/š‘‘š‘„=((š‘„ āˆ’ 1)^ā€² (š‘„ āˆ’ 2) āˆ’ (š‘„ āˆ’ 2)^ā€² (š‘„ āˆ’ 1))/(š‘„ āˆ’ 2)^2 Using Quotient Rule As (š‘¢/š‘£)^ā€²=(š‘¢^ā€² š‘£ āˆ’怖 š‘£ć€—^ā€² š‘¢)/š‘£^2 š‘‘š‘¦/š‘‘š‘„=(1 (š‘„ āˆ’ 2) āˆ’1 (š‘„ āˆ’ 1))/(š‘„ āˆ’ 2)^2 š‘‘š‘¦/š‘‘š‘„=(š‘„ āˆ’ 2 āˆ’ š‘„ + 1)/(š‘„ āˆ’ 2)^2 So, š‘‘š‘¦/š‘‘š‘„=(āˆ’ 1)/(š‘„ āˆ’ 2)^2 Putting š‘„=10 (š‘‘š‘¦/š‘‘š‘„)_(š‘„ = 10)=(āˆ’ 1)/(10 āˆ’ 2)^2 =(āˆ’ 1)/怖 8怗^2 =(āˆ’ 1)/64 Hence Slope of a tangent at š‘„ = 10 is (āˆ’šŸ)/šŸ”šŸ’

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