<s>[INST] error in solutions? MIT single variable calculus - Simplifying formula for tangent line Question 1C-4 d) located here states:

Write an equation for the tangent line for the following questions:
  d) $f(x) = \frac{1}{\sqrt x} $ at $x = a$.

By my workings:
$f'(a) = (\frac{-1}{2})a^\frac{-3}{2}$
Therefore, the equation of the tangent line at $x = a$ is:
$y = (\frac{-1}{2})a^\frac{-3}{2}[x-a] + \frac{1}{\sqrt a}$
My questions:


*

*The solutions pdf located here states that the formula for the tangent line at $x = a$ is 
$y = (\frac{-1}{2})a^\frac{3}{2}[x-a] + \frac{1}{\sqrt a}$. The difference to mine is that I have a negative exponent to the first $a$. Is the solution correct?

*The solutions say that the equation for the tangent line at $x = a$ simplifies to $y = -a^\frac{-3}{2}x + (\frac{3}{2})a^\frac{-1}{2}$. Could you show me how to get to this simplified solution, explicitly stating each step and  which rules have been applied to get to it?


Thank you.
 [/INST] For the simplifcation:
$$ y = -\frac{1}{2}a^{-\frac{3}{2}}(x-a)+\frac{1}{\sqrt{a}} $$
$$ y = \frac{1}{2}a^{-\frac{3}{2}}\cdot a -\frac{1}{2}a^{-\frac{3}{2}} \cdot x + a^{-\frac{1}{2}} \\ \text{(multiplication, and definition of power and square root)} $$
$$ y = \frac{1}{2}a^{-\frac{1}{2}} + a^{-\frac{1}{2}} - \frac{1}{2}a^{-\frac{3}{2}}\cdot x \\ \text{(reordering)} $$
$$ y = \frac{3}{2}a^{-\frac{1}{2}} - a^{-\frac{3}{2}}x \\ \text{(summation)} $$
As was to be shown.
(I am not too familiar with the syntax, I hope it is clear though)
</s>