Answer:
[tex]h'(t)=-\dfrac{14}{5t^3}+\dfrac{1}{t^{2}}[/tex]
Step-by-step explanation:
Consider the given function is
[tex]h(t)=\dfrac{(\dfrac{7}{t^2})}{5}-\dfrac{(\dfrac{2}{t})}{2}[/tex]
On simplification we get
[tex]h(t)=\dfrac{7}{5t^2}-\dfrac{2}{2t}[/tex]
[tex]h(t)=\dfrac{7}{5}(t^{-2})-(t^{-1})[/tex]
Differentiated with respect to t.
[tex]h'(t)=\dfrac{7}{5}(-2)t^{-3}-(-1)t^{-2}[/tex]
[tex]h'(t)=-\dfrac{14}{5t^3}+\dfrac{1}{t^{2}}[/tex]
Therefore, [tex]h'(t)=-\dfrac{14}{5t^3}+\dfrac{1}{t^{2}}[/tex].
