[tex]f(t)=6.5t^{-10}[/tex]
then we have a new function:
[tex]g(t)\text{ = }ln(6.5t^{-10}\text{)}[/tex]
its derivative is: the derivative of the outer function (natural logarithm) times the derivative of the inner function 6.5t^-10
that is:
[tex]\frac{dg(t)}{d(t)}\text{ = }\frac{1}{6.5t^{-10}}\frac{d}{d\text{ t}}\text{( }6.5t^{-10})^{\square}[/tex]
the above because the derivative of the function ln is 1 / x , where in this case x = 6.5t^-10.
The above equation is equivalent to:
[tex]\frac{dg(t)}{d(t)}\text{ = }\frac{1}{6.5t^{-10}}\text{( -10 x}6.5t^{-11})^{\square}=\text{ }\frac{1}{6.5t^{-10}}\text{( -}65t^{-11})=-10t^{-1}[/tex]
So the correct answer is:
[tex]\frac{dg(t)}{d(t)}\text{ = }\frac{d}{dt}\text{ }\ln (f(t))\text{= }-10t^{-1}[/tex]
