Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.4 question 3

need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.4 question 3

Answers (1)

Answer:

-\left(\frac{y}{x}\right)^{\frac{1}{3}}

Hint:

Use the differentiation formula of \left(x^{n}\right)

i.e. \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}

Given:

x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}

Solution:

Differentiate the given equation w.r.t x

\frac{d}{d x}\left(x^{\frac{2}{3}}+y^{\frac{2}{3}}\right)=\frac{d}{d x}\left(a^{\frac{2}{3}}\right)

\frac{d}{d x}\left(x^{\frac{2}{3}}\right)+\frac{d}{d x}\left(y^{\frac{2}{3}}\right)=0 \quad\left[\because \frac{d(\operatorname{cons} \tan t)}{d x}=0\right]

\frac{2}{3}(x)^{\frac{2}{3}-1}+\frac{d\left(y^{\frac{2}{3}}\right)}{d y} \times \frac{d y}{d x}=0 \quad\left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]

\frac{2}{3}(x)^{\frac{-1}{3}}+\frac{2}{3}(y)^{\frac{2}{3}-1} \frac{d y}{d x}=0

\frac{2}{3}(x)^{\frac{-1}{3}}+\frac{2}{3}(y)^{\frac{-1}{3}} \frac{d y}{d x}=0

\frac{d y}{d x}=\frac{\frac{-2}{3}(x)^{\frac{-1}{3}}}{\frac{2}{3}(y)^{\frac{-1}{3}}}=\frac{-(x)^{\frac{-1}{3}}}{(y)^{\frac{-1}{3}}}

\frac{d y}{d x}=-\frac{(y)^{\frac{1}{3}}}{(x)^{\frac{1}{3}}} \quad\left[\because(x)^{-n}=\frac{1}{(x)^{n}}\right]

\frac{d y}{d x}=-\left(\frac{y}{x}\right)^{\frac{1}{3}}

Hence  \frac{d y}{d x}=-\left(\frac{y}{x}\right)^{\frac{1}{3}}is required answer

 

 

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 Answer Sheets: CIC tells board to charge photocopying fee as per RTI rules, amend circular
anam-khan

CBSE Class 12 Supplementary Result 2026: Scanned copy, verification, re-evaluation dates announced
anam-khan

SC asks CBSE to protect students affected by OSM evaluation in admissions

Latest Question