Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • explain solution RD Sharma class 12 chapter Differentiation exercise 10.2 question 66 maths

explain solution RD Sharma class 12 chapter Differentiation exercise 10.2 question 66 maths

Answers (1)

best_answer

Answer: Proved

Hint: you must know the rules of derivative of logarithm functions.

Given:  \mathrm{y}=(\mathrm{x}-1) \log (x-1)-(x+1) \log (x+1)

Prove: \frac{d y}{d x}=\log \left(\frac{x-1}{1+x}\right)

Solution:

Let  \mathrm{y}=(\mathrm{x}-1) \log (x-1)-(x+1) \log (x+1)

Differentiate with respect to x, use product rule

\frac{d y}{d x}=\frac{d}{d x}[(\mathrm{x}-1) \log (x-1)-(x+1) \log (x+1)]

\frac{d y}{d x}=\left[(x-1) \times \frac{1}{(x-1)}+\log (x-1)\right]-\left[(x+1) \times \frac{1}{(x+1)}+\log (x+1)\right]   [ use product rule]         

\frac{d y}{d x}=[1+\log (x-1)]-[1+\log (x+1)]

\frac{d y}{d x}=\log \left(\frac{x-1}{1+x}\right)

∴ Proved

Posted by

infoexpert26

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Board exams twice a year from 2025, no plan for semesters: MoE
anam-khan

CBSE Class 12 board exams 2025 to have 50% competency-based questions; no change in 10th
anam-khan

CBSE Class 12 board exams 2024 over; expected result date, trends of last 10 years

Latest Question