We have the following curve:
[tex]y=\frac{\left | x \right |}{5-x^2}[/tex]
So we need to find an equation of the tangent line to this curve at the point [tex](2,2)[/tex]. So let's find out if this point, in fact, belongs to the curve:
[tex]If \ x=2 \\ \\ y=\frac{\left | 2 \right |}{5-2^2}=\frac{\left | 2 \right |}{5-4}=2[/tex].
We also know that:
[tex]\left | x \right |= \left \{ {{x \ if \ x \ \geq 0} \atop {-x \ if \ x\ \textless \ 0}} \right.[/tex]
Given that the point is:
[tex] (2,2)\ that \ is: \\ x=2\ \textgreater \ 0[/tex]
Then we will say that:
[tex]\left | x \right |=x[/tex]
Therefore:
[tex]y=\frac{x}{5-x^2}[/tex]
Computing the derivative:
[tex]y=\frac{x}{5-x^2} \\ \frac{dy}{dx}= \frac{(1)(5-x^2)-(-2x)x}{(5-x^2)^2}=\frac{5+x^2}{(5-x^2)^2}[/tex]
So the derivative solved for [tex]x=2[/tex] is in fact the slope of the line at the point [tex](2,2)[/tex], then:
[tex]\frac{dy}{dx} |_{x=2}=\frac{5+x^2}{(5-x^2)^2}|_{x=2}=\frac{5+2^2}{(5-2^2)^2}=9 \\ \\ \therefore m=\frac{dy}{dx} |_{x=2}=9[/tex]
Finally, the tangent line is:
[tex]y-y_1=m(x-x_1) \\ \therefore y-2=9(x-2) \\ \therefore \boxed{y=9x-16}[/tex]
This is shown in the figure below.
