CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question
Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
If $\text{y}=\text{x}\sin^{-1}\text{x}+\sqrt{1-\text{x}^2},$ prove that $\frac{\text{dy}}{\text{dx}}=\sin^{-1}\text{x}$
✓
Answer
We have, $\text{y}=\text{x}\sin^{-1}\text{x}+\sqrt{1-\text{x}^2}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big[\text{x}\sin^{-1}\text{x}+\sqrt{1-\text{x}^2}\Big]$
$=\frac{\text{d}}{\text{dx}}\big(\text{x}\sin^{-1}\text{x}\big)+\frac{\text{d}}{\text{dx}}\big(\sqrt{1-\text{x}^2}\big)$
$=\Big[\text{x}\frac{\text{d}}{\text{dx}}\sin^{-1}\text{x}+\sin^{-1}\text{x}\frac{\text{d}}{\text{dx}}(\text{x})\Big]+\frac{1}{2\sqrt{1-\text{x}^2}}\frac{\text{d}}{\text{dx}}(1-\text{x}^2)$
$=\Big[\frac{\text{x}}{\sqrt{1-\text{x}^2}}+\sin^{-1}\text{x}\Big]-\frac{2\text{x}}{2\sqrt{1-\text{x}^2}}$
$=\frac{\text{x}}{\sqrt{1-\text{x}^2}}+\sin^{-1}\text{x}-\frac{\text{x}}{\sqrt{1-\text{x}^2}}$
$=\sin^{-1}\text{x}$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.