Search for: Mathematicsddxlogexx−2x+23/4 is ddxlogexx−2x+23/4 is A1Bx2+1x2−4Cx2−1x2−4Dexx2−1x2−4 Congratulations you have unlocked a coupon code of 10% INFY10 Check It Out Fill Out the Form for Expert Academic Guidance!l Grade ---Class 6Class 7Class 8Class 9Class 10Class 11Class 12 Target Exam JEENEETCBSE +91 Preferred time slot for the call ---9 am10 am11 am12 pm1 pm2 pm3 pm4 pm5 pm6 pm7 pm8pm9 pm10pm Please indicate your interest Live ClassesBooksTest SeriesSelf Learning Language ---EnglishHindiMarathiTamilTeluguMalayalam Are you a Sri Chaitanya student? NoYes Verify OTP Code (required) I agree to the terms and conditions and privacy policy. Solution:let y=logexx−2x+23/4=logex+logx−2x+23/4⇒y=x+34[log(x−2)−log(x+2)]On differentiating w.r.t. x, we getdydx=ddxx+34{log(x−2)−log(x+2)}=1+341x−2−1x+2=1+3x2−4⇒dydx=x2−1x2−4 Related content Area of Square Area of Isosceles Triangle Pythagoras Theorem Triangle Formulae Perimeter of Triangle Formula Area Formulae Volume of Cone Formula Matrices and Determinants_mathematics Critical Points Solved Examples Type of relations_mathematics