Ex 6.3,4 - Chapter 6 Class 12 Application of Derivatives (Term 1)
Last updated at April 14, 2021 by Teachoo
Transcript
Ex 6.3, 4 Find the slope of the tangent to the curve π¦=π₯^3β3π₯+2 at the point whose π₯βcoordinate is 3 π¦=π₯^3β3π₯+2 We know that slope of tangent =ππ¦/ππ₯ ππ¦/ππ₯=3π₯^2β3 Since π₯βcoordinate is 3 Putting π₯=3 in (1) γππ¦/ππ₯βγ_(π₯ = 3)=3(3)^2β3 =3 Γ9β3 =27β3 =24 Hence slope of tangent is 24
Ex 6.3
Ex 6.3, 1
Ex 6.3,2
Ex 6.3,3
Ex 6.3,4 You are here
Ex 6.3, 5 Important
Ex 6.3,6
Ex 6.3,7
Ex 6.3,8
Ex 6.3,9 Important
Ex 6.3,10
Ex 6.3,11 Important
Ex 6.3,12
Ex 6.3,13
Ex 6.3,14 Important
Ex 6.3,15 Important
Ex 6.3,16
Ex 6.3,17
Ex 6.3,18 Important
Ex 6.3,19
Ex 6.3,20
Ex 6.3,21
Ex 6.3,22
Ex 6.3,23 Important
Ex 6.3,24 Important
Ex 6.3,25
Ex 6.3,26 Important
Ex 6.3,27
Chapter 6 Class 12 Application of Derivatives (Term 1)
Serial order wise
About the Author
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.