differentiate w.r.t x: xsin¯¹x÷ (1-x²) …
CBSE, JEE, NEET, CUET
Question Bank, Mock Tests, Exam Papers
NCERT Solutions, Sample Papers, Notes, Videos
Posted by Yatin Jindal 6 years, 9 months ago
- 1 answers
Related Questions
Posted by Anupam Bind Anupam Bind 3 weeks, 6 days ago
- 0 answers
Posted by Dhruv Kumar 1 month, 2 weeks ago
- 1 answers
Posted by Arpit Sahu 4 weeks, 2 days ago
- 0 answers
Posted by Suprith A 1 month ago
- 0 answers
Posted by Pinky Kumari 2 weeks, 3 days ago
- 0 answers
Posted by Durgesh Kuntal Nishantkuntal 1 week, 3 days ago
- 0 answers
myCBSEguide
Trusted by 1 Crore+ Students
Test Generator
Create papers online. It's FREE.
CUET Mock Tests
75,000+ questions to practice only on myCBSEguide app
Rashmi Bajpayee 6 years, 9 months ago
Let {tex}y = {{x{{\sin }^{ - 1}}x} \over {1 - {x^2}}}{/tex}
=> {tex}{{dy} \over {dx}} = {{\left( {1 - {x^2}} \right){d \over {dx}}\left( {x{{\sin }^{ - 1}}x} \right) - x{{\sin }^{ - 1}}x{d \over {dx}}\left( {1 - {x^2}} \right)} \over {{{\left( {1 - {x^2}} \right)}^2}}}{/tex}
=> {tex}{{dy} \over {dx}} = {{\left( {1 - {x^2}} \right)\left[ {x.{1 \over {\sqrt {1 - {x^2}} }} + {{\sin }^{ - 1}}x.1} \right] + 2{x^2}{{\sin }^{ - 1}}x} \over {{{\left( {1 - {x^2}} \right)}^2}}}{/tex}
=> {tex}{{dy} \over {dx}} = {{x\sqrt {1 - {x^2}} + \left( {1 - {x^2}} \right){{\sin }^{ - 1}}x + 2{x^2}{{\sin }^{ - 1}}x} \over {{{\left( {1 - {x^2}} \right)}^2}}}{/tex}
=> {tex}{{dy} \over {dx}} = {{x\sqrt {1 - {x^2}} + {{\sin }^{ - 1}}x - {x^2}{{\sin }^{ - 1}}x + 2{x^2}{{\sin }^{ - 1}}x} \over {{{\left( {1 - {x^2}} \right)}^2}}}{/tex}
=> {tex}{{dy} \over {dx}} = {{x\sqrt {1 - {x^2}} + {{\sin }^{ - 1}}x + {x^2}{{\sin }^{ - 1}}x} \over {{{\left( {1 - {x^2}} \right)}^2}}}{/tex}
=> {tex}{{dy} \over {dx}} = {{x\sqrt {1 - {x^2}} + {{\sin }^{ - 1}}x\left( {1 + {x^2}} \right)} \over {{{\left( {1 - {x^2}} \right)}^2}}}{/tex}
0Thank You