Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • provide solution for RD Sharma maths class 12 chapter Differentiation exercise 10.2 question 61

provide solution for RD Sharma maths class 12 chapter Differentiation exercise  10.2 question 61

Answers (1)

Answer:   Proved

Hint: you must know the rule of solving logarithm functions.

Given:   \mathrm{y}=\log \left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)

Prove   \frac{d y}{d x}=\frac{x-1}{2 x(x+1)}

Solution:

Let  \mathrm{y}=\log \left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)

Differentiate with respect to x

\begin{aligned} &\frac{d y}{d x}=\frac{1}{\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)} \cdot \frac{d}{d x}\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right) \\\\ &\frac{d y}{d x}=\frac{\sqrt{x}}{x+1}\left(\frac{1}{2 \sqrt{x}}-\frac{1}{2 x \sqrt{x}}\right) \end{aligned}

\begin{aligned} &\frac{d y}{d x}=\frac{1}{2} \frac{\sqrt{x}}{(x+1)}\left(\frac{x-1}{x \sqrt{x}}\right) \\\\ &\frac{d y}{d x}=\frac{x-1}{2 x(x+1)} \end{aligned}

∴ 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 Single Girl Child Scholarship 2024: Registration extended till January 10; eligibility criteria, amount
anam-khan

CBSE wants international boards reined in; letter to education ministry seeks directions for AIU
anam-khan

PM Modi's Pariksha Pe Charcha 2025 to be held next month, registrations open till January 14

Latest Question