## University Calculus: Early Transcendentals (3rd Edition)

Differentiate the RHS and see if it equals$\quad(\sin^{-1}x)^{2}$ $\displaystyle \frac{d}{dx}[x\sin^{-1}x]=(1)(\sin^{-1}x)^{2}+x\cdot 2\sin^{-1}x\cdot\frac{1}{\sqrt{1-x^{2}}}\cdot(-2x)$ $=(\displaystyle \sin^{-1}x)^{2}-\frac{4x\sin^{-1}x}{\sqrt{1-x^{2}}}$ $\displaystyle \frac{d}{dx}[2x]=2$ $\displaystyle \frac{d}{dx}[2\sqrt{1-x^{2}}\cdot\sin^{-1}x]=2\cdot\frac{-1}{\sqrt{1-x^{2}}}\cdot(-2x)\cdot\sin^{-1}x+2\sqrt{1-x^{2}}\cdot\frac{1}{\sqrt{1-x^{2}}}]$ $=\displaystyle \frac{4x\sin^{-1}x}{\sqrt{1-x^{2}}}+2$ $\displaystyle \frac{d}{dx}[C]=0$ $\displaystyle \frac{d}{dx}[RHS]=(\sin^{-1}x)^{2}-\frac{4x\sin^{-1}x}{\sqrt{1-x^{2}}}-2+\frac{4x\sin^{-1}x}{\sqrt{1-x^{2}}}+2$ $=(\sin^{-1}x)^{2}$ which verifies the formula as needed.