#### Answer

$y=x$

#### Work Step by Step

The slope of the tangent at x=a
is the derivative of f(x), calculated in x=a.
The tangent line has slope $f^{\prime}(a)$, and passes through $(a,f(a)).\\\\$
The point-slope equation is
$y-f(a)=f^{\prime}(a)(x-a)$
We find $f'(x):$
$f(x)=x(x^{2}-4x+5)^{4}=u(x)\cdot v(x)\qquad $(... use the product rule)
$u(x)=x\qquad u^{\prime}(x)=1$
$v(x)=(x^{2}-4x+5)^{4}\qquad $(... we use the chain rule)
$v^{\prime}(x)=4(x^{2}-4x+5)^{3}\cdot(x^{2}-4x+5)^{\prime}\\=4(x^{2}-4x+5)^{3}\cdot(2x-4)$
$f^{\prime}(x)=u(x)\cdot v^{\prime}(x)+v(x)\cdot u^{\prime}(x)$.
$=x\cdot 4(x^{2}-4x+5)^{3}\cdot(2x-4) +1\cdot(x^{2}-4x+5)^{4}$
$=(x^{2}-4x+5)^{3}\cdot [4x(2x-4)+(x^{2}-4x+5)]$
$\color{blue}{f^{\prime}(x)=(x^{2}-4x+5)^{3}(9x^{2}-20x+5)}$
$f(2)=x(4-4(2)+5)^{4}=2(1)^{4}=2$
$f^{\prime}(2)=(4-4(2)+5)^{3}(9(4)-20(2)+5)=(1)^{3}(1)=1$
Point-slope form for the line passing through $(2,2),$ with slope $1:$
$y-2=1(x-2)$
... solve for y (to transform to slope-intercept form)...
$y-2=x-2$
$y=x-2+2$
$y=x$