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 R.D. Sharma class 12 chapter 10 Differentiation exercise 10.3 question 6 maths textbook solution.

Provide solution for R.D. Sharma  class 12 chapter 10 Differentiation exercise 10.3 question 6 maths textbook solution.

Answers (1)

Answer :  \frac{d y}{d x}=\frac{a}{a^{2}+x^{2}}
Hint:


Given:

\sin ^{-1}\left\{\frac{\mathrm{x}}{\sqrt{\mathrm{x}^{2}+\mathrm{a}^{2}}}\right\}

Solution:

\begin {array} {ll} Let \ y=\sin ^{-1}\left\{-\frac{x}{\sqrt{x^{2}+a^{2}}}\right\}\\\\ Let \ \mathrm{x}=\operatorname{atan} \theta\\\\ Now,\\ \mathrm{y}=\sin ^{-1}\left\{\frac{\operatorname{atan} \theta}{\sqrt{a^{2} \tan ^{2} \theta+a^{2}}}\right\}\\\\ \mathrm{y}=\sin ^{-1}\left\{\frac{\text { atsan } \theta}{\sqrt{a^{2}\left(\tan ^{2} \theta+1\right)}}\right\}\\\\\ \mathrm{y}=\sin ^{-1}\left\{\frac{\operatorname{atan} \theta}{\mathrm{a}\left(\tan ^{2} \theta+1\right)}\right\}\\\\ \left.\mathrm{y}=\sin ^{-1} \int \frac{\operatorname{atan} \theta}{\mathrm{a} \sqrt{\sec ^{2} \theta}}\right\} \end{array}

\begin{array}{l} \text { Using }:\left\{\tan \theta=\frac{\sin \theta}{\cos \theta}, \cos \theta=\frac{1}{\sec \theta}\right\}\\\\ \mathrm{y}=\sin ^{-1}\{\sin \theta\}\\\\ \mathrm{y}=\theta \quad \ \ \ \ \ \ \ \ \ \therefore\left\{\sin ^{-1} \theta(\sin \theta)=\theta\right\} \quad \text { if } \theta \varepsilon\left[\frac{-\pi}{2}, \frac{\mathrm{\pi}}{2}\right]\\\\ \mathrm{x}=\operatorname{atan} \theta\\\\ \mathrm{x}=\operatorname{atan} \theta\\\\ \theta=\tan ^{-1}\left(\frac{x}{a}\right)\\\\ y=\tan ^{-1}\left(\frac{x}{a}\right)\\\\ \end{array}

differentiating with respects to x, we get

\begin{array}{l} \frac{d y}{d x}=\frac{d}{d x}\left(\tan ^{-1}\left(\frac{x}{a}\right)\right) \\\\ \frac{\partial}{\partial x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^{2}} \\\\ \frac{d y}{d x}=\left(\frac{1}{1+\left(\frac{x}{a}\right)^{2}}\right) \times \frac{1}{a} \\\\ \end{array}

\begin{array}{l} \frac{d y}{d x}=\left(\frac{1}{1+\frac{x^{2}}{a^{2}}}\right) \times \frac{1}{a} \\\\ \frac{d y}{d x}=\left(\frac{1}{\frac{a^{2}+x^{2}}{a^2}}\right) \times \frac{1}{a} \\\\ \frac{d y}{d x}=\frac{a^{2}}{a^{2}+x^{2}} \times \frac{1}{a} \\\\ \frac{d y}{d x}=\frac{a}{a^{2}+x^{2}} \end{array}

Posted by

infoexpert22

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 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