Answer:
The first derivative of [tex]r(t) = 5\cdot t^{-2}[/tex] (r(t)=5*t^{-2}) with respect to t is [tex]r'(t) = -10\cdot t^{-3}[/tex] (r'(t) = -10*t^{-3}).
Step-by-step explanation:
Let be [tex]r(t) = \frac{5}{t^{2}}[/tex], which can be rewritten as [tex]r(t) = 5\cdot t^{-2}[/tex]. The rule of differentiation for a potential function multiplied by a constant is:
[tex]\frac{d}{dt}(c \cdot t^{n}) = n\cdot c \cdot t^{n-1}[/tex], [tex]\forall \,n\neq 0[/tex]
Then,
[tex]r'(t) = (-2)\cdot 5\cdot t^{-3}[/tex]
[tex]r'(t) = -10\cdot t^{-3}[/tex] (r'(t) = -10*t^{-3})
The first derivative of [tex]r(t) = 5\cdot t^{-2}[/tex] (r(t)=5*t^{-2}) with respect to t is [tex]r'(t) = -10\cdot t^{-3}[/tex] (r'(t) = -10*t^{-3}).