
Slope of Tangent
The gravitational potential energy of the MIR space station, in joules, at a distance of x kilometers from the centre of the Earth, is described by the equation
$\displaystyle E = \frac{(5.38)(10)^{19}}{x}$. Determine the slope of the tangent to the graph at the given value of x.
$\displaystyle x = (6.5)(10)^6 m$
My answer:
$\displaystyle x = (6.5)(10)^6 m = 6500 km$
$\displaystyle m = \frac{E(6500 + 0.001)  E(6500)}{0.001}$
$\displaystyle m = \frac{(\frac{(5.38)(10)^{19}}{(6500 + 0.001)}) (\frac{(5.38)(10)^{19}}{(6500)})}{0.001}$
$\displaystyle m = 1.273(10)^{12}$
Is this correct?

Yes your anwers is correct.
Your formula can be rewritten as:
$\displaystyle E = (5.38)(10)^{19}* \frac{1}{x}$
Which has derivative:
$\displaystyle E' = (5.38)(10)^{19}* \frac{1}{x^2}$
Proof see: Calculus with polynomials  Wikipedia, the free encyclopedia
With your x, this gives $\displaystyle 1.27337278 \times 10^{12}$