Answer
The equation for the tangent line is:
$$y=\frac{7}{2}x-\frac{3}{2}.$$
Work Step by Step
The equation for the tangent line at $x_0$ is:
$$y-f(x_0)=f'(x_0)(x-x_0)$$
Here is $x_0=1$ and $f(x)=x^2\sqrt{5-x^2}.$
$$f'(x)=(x^2\sqrt{5-x^2})'=(x^2)'\sqrt{5-x^2}+x^2(\sqrt{5-x^2})'=
2x\sqrt{5-x^2}+x^2\frac{1}{2\sqrt{5-x^2}}(5-x^2)'=
2x\sqrt{5-x^2}+x^2\frac{1}{2\sqrt{5-x^2}}\cdot(-2x)=
2x\sqrt{5-x^2}-\frac{x^3}{\sqrt{5-x^2}}=\frac{2x(5-x^2)-x^3}{\sqrt{5-x^2}}=\frac{10x-3x^3}{\sqrt{5-x^2}}$$
For $x=1$ we have:
$$f(1)=1^2\sqrt{5-1^2}=1\cdot\sqrt{4}=2$$
$$f'(1)=\frac{10\cdot1-3\cdot1^3}{\sqrt{5-1^2}}=\frac{7}{2}$$
So, the equation for the tangent line is:
$$y-2=\frac{7}{2}(x-1)\Rightarrow y=\frac{7}{2}x-\frac{7}{2}+2\Rightarrow y=\frac{7}{2}x-\frac{3}{2}$$
