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 15

need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.4 question 15

Answers (1)

Answer:

2 \frac{d y}{d x}+y^{3}=0

Hint:

Use product rule and \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}

Given:

x y^{2}=1

Solution:

Differentiate \left[x y^{2}=1\right]  w.r.t x

\frac{d\left(x y^{2}\right)}{d x}=\frac{d(1)}{d x}

x \frac{d y^{2}}{d x}+y^{2} \frac{d x}{d x}=0                \left[\because \frac{d(\operatorname{cons} \tan t)}{d x}=0\right]    [Product rule \frac{d(u v)}{d x}=u \cdot \frac{d v}{d x}+v \cdot \frac{d u}{d x}]

x \frac{d y^{2}}{d y} \times \frac{d y}{d x}+y^{2}=0

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

2 x y \frac{d y}{d x}+y^{2}=0

\begin{aligned} &\frac{d y}{d x}=-\frac{y^{2}}{2 x y}=\frac{-y}{2 x} \\ &2 \frac{d y}{d x}=-\frac{y}{x} \end{aligned}

2 \frac{d y}{d x}=-\frac{y}{\left(\frac{1}{y^{2}}\right)}                    \left[\begin{array}{c} x y^{2}=1 \\ x=\frac{1}{y^{2}} \end{array}\right]

2 \frac{d y}{d x}+y^{3}=0

Hence, if  x y^{2}=1  then 2 \frac{d y}{d x}+y^{3}=0  is the required answer

Hence 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 10th, 12th practical exams 2025 begin; instructions for students, schools
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

Latest Question