[tex] \frac{d}{dx}(x\sqrt{16-x^2}) = \frac{d}{dx}(x) \cdot \sqrt{16-x^2} + \frac{d}{dx}(\sqrt{16-x^2}) \cdot x[/tex] by the product rule.
Notice, [tex] \frac{d}{dx}(\sqrt{16-x^2}) = \frac{d}{dx}(({16-x^2})^ \frac{1}{2}) = \frac{1}{2}(16-x^2 )^-0.5 \cdot \frac{d}{dx} (16-x^2)[/tex], by the chain rule so we have
[tex]\frac{d}{dx}(x\sqrt{16-x^2}) = 1 \cdot \sqrt{16-x^2} -x(16-x^2)^{- \frac{1}{2}} \cdot x[/tex] which simplifies to
[tex]\frac{d}{dx}(x\sqrt{16-x^2}) = \sqrt{16-x^2} -x^2(16-x^2)^{- \frac{1}{2}}[/tex]
