Solutions

a) Let $g(x)=\sqrt{x}$ and $h(x)=1-x^2$. Hence,

$$\begin{align*} f'(x)&=(g\circ h)'(x) \\ &=g'(h(x))\cdot h'(x) \\ &=\frac{1}{2\sqrt{1-x^2}}\cdot (-2x) \\ &=\frac{-x}{\sqrt{1-x^2}}. \end{align*}$$

b) The tangent line of $f$ at $(a,\sqrt{1-a^2})$ is defined as

$$y(x)=\frac{-a}{\sqrt{1-a^2}}(x-a)+\sqrt{1-a^2}.$$

It intersects with $f$ when

$$y(x)=f(x)\implies \frac{-a}{\sqrt{1-a^2}}(x-a)+\sqrt{1-a^2}=\sqrt{1-x^2}.$$

Then

$$\begin{align*} &\implies \frac{1-a^2-2a(x-a)+a^2(x-a)^2}{1-a^2}=1-x^2 \\ &\implies 1-a^2-2ax+2a^2+a^2(x^2-2ax+a^2)=1-x^2-a^2+a^2x^2 \\ &\implies 1-2ax+a^2x^2=1-x^2-a^2+a^2x^2 \\ &\implies x^2-2ax+a^2=0 \\ &\implies (x-a)^2=0 \\ &\implies x=a. \end{align*}$$

Thus, $y$ intersects $f$ only once at $x=a$.