+91-8076753736 support@wiredfaculty.com
logo
logo

A PHP Error was encountered

Severity: Notice

Message: Undefined variable: temp_qds

Filename: Questions_Page/Ncert_Question.php

Line Number: 320

Backtrace:

File: /home/wiredfa1/public_html/application/views/final/Questions_Page/Ncert_Question.php
Line: 320
Function: _error_handler

File: /home/wiredfa1/public_html/application/controllers/Home.php
Line: 235
Function: view

File: /home/wiredfa1/public_html/index.php
Line: 315
Function: require_once

Sponsor Area

Continuity And Differentiability

Question
CBSEENMA12035215
Wired Faculty App

Differentiate space tan to the power of negative 1 end exponent open square brackets fraction numerator square root of 1 plus straight a squared space straight x squared end root minus 1 over denominator straight a space straight x end fraction close square brackets space straight w. straight r. straight t. straight x.

Solution
Let space straight y equals tan to the power of negative 1 end exponent open square brackets fraction numerator square root of 1 plus straight a squared space straight x squared end root minus 1 over denominator straight a space straight x end fraction close square brackets semicolon space Put space straight a space straight x equals tan space straight theta space space space space space space space space space equals tan to the power of negative 1 end exponent open square brackets fraction numerator square root of 1 plus tan squared space straight theta end root minus 1 over denominator tan space straight theta end fraction close square brackets equals tan to the power of negative 1 end exponent open square brackets fraction numerator sec space straight theta minus 1 over denominator tan space straight theta end fraction close square brackets space space space space space space space space space equals tan to the power of negative 1 end exponent open square brackets fraction numerator begin display style fraction numerator 1 over denominator cos space straight theta end fraction end style minus 1 over denominator begin display style fraction numerator sin space straight theta over denominator cos space straight theta end fraction end style end fraction close square brackets equals equals tan to the power of negative 1 end exponent open square brackets fraction numerator 1 minus cos space straight theta over denominator sin space straight theta end fraction close square brackets space space space space space space space space space equals tan to the power of negative 1 end exponent open square brackets fraction numerator 2 space sin squared begin display style straight theta over 2 end style over denominator 2 space sin begin display style straight theta over 2 end style cos begin display style straight theta over 2 end style end fraction close square brackets equals tan to the power of negative 1 end exponent open square brackets tan space straight theta over 2 close square brackets equals straight theta over 2 space space space space space space space straight y equals 1 half tan to the power of negative 1 end exponent straight x therefore space dy over dx equals 1 half. fraction numerator straight a over denominator 1 plus left parenthesis straight a space straight x right parenthesis squared end fraction equals fraction numerator straight a over denominator 2 left parenthesis 1 plus straight a squared space straight x squared right parenthesis end fraction

Some More Questions From Continuity and Differentiability Chapter

Mock Test Series

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation