Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferentiation4 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, $\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big[\text{x}\sin^{-1}\text{x}+\sqrt{1-\text{x}^2}\Big]$ $=\frac{\text{d}}{\text{dx}}\big(\text{x}\sin^{-1}\text{x}\big)+\frac{\text{d}}{\text{dx}}\big(\sqrt{1-\text{x}^2}\big)$ $=\Big[\text{x}\frac{\text{d}}{\text{dx}}\sin^{-1}\text{x}+\sin^{-1}\text{x}\frac{\text{d}}{\text{dx}}(\text{x})\Big]+\frac{1}{2\sqrt{1-\text{x}^2}}\frac{\text{d}}{\text{dx}}(1-\text{x}^2)$ $=\Big[\frac{\text{x}}{\sqrt{1-\text{x}^2}}+\sin^{-1}\text{x}\Big]-\frac{2\text{x}}{2\sqrt{1-\text{x}^2}}$ $=\frac{\text{x}}{\sqrt{1-\text{x}^2}}+\sin^{-1}\text{x}-\frac{\text{x}}{\sqrt{1-\text{x}^2}}$ $=\sin^{-1}\text{x}$
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