Find the slope of the tangent line to y=x^5-3x^-3 at x=-2

The slope of a tangent drawn on a curve y = f(x) at a
point where x = a is given by the f'(a) where f'(x) is the first derivative of y =
f(x).

Here y = f(x) is
y=x^5-3x^-3

f'(x) = (x^5 -
3x^-3)'

= 5x^4 -
3*(-3)*x^(-3-1)

= 5x^4 +
9*x^-4

At the point where x =
-2

f'(-2) = 5*(-2)^4 +
9*(-2)^-4

= 80 + 9/16

=
1289/16

The slope of the required tangent is
1289/16