Answer
(a) The derivative of
$$
y=\sqrt{3+x^{2}}
$$
is
$$
d y=\frac{1}{2}\left(3+x^{2}\right)^{-1 / 2}(2 x) d x=\frac{x}{\sqrt{3+x^{2}}} d x
$$
(b) When $$ x=1 \,\,\,\ \text{and} \,\,\,\,\, d x=-0.1$$,
$$
d y=\frac{1}{\sqrt{3+1^{2}}}(-0.1)=\frac{1}{2}(-0.1)=-0.05
$$
Work Step by Step
(a) The derivative of
$$
y=\sqrt{3+x^{2}}
$$
is
$$
d y=\frac{1}{2}\left(3+x^{2}\right)^{-1 / 2}(2 x) d x=\frac{x}{\sqrt{3+x^{2}}} d x
$$
(b) When $$ x=1 \,\,\,\ \text{and} \,\,\,\,\, d x=-0.1$$,
$$
d y=\frac{1}{\sqrt{3+1^{2}}}(-0.1)=\frac{1}{2}(-0.1)=-0.05
$$
