We have to find

\((\frac{d}{dt})(5\ln(\sin(t)+8))\)

Note that

\((\frac{d}{dt})(5\ln(\sin(t)+8))=5\frac{d}{dt}(\ln(\sin(t)+8)) [∵(c⋅f) ]\)

\(=5\frac{d}{dt}(\ln u), =5\frac{d}{dt}(\ln (u))(\frac{d}{dt})(\sin(t)+8) =5*(\frac{1}{u})\cos(t) =\frac{5\cos(t)}{\sin(t)+8}/\)

Hence we get

\(\frac{d}{dt}(5\ln(t)+8))=\frac{5\cos(t)}{\sin(t)+8}\)

\((\frac{d}{dt})(5\ln(\sin(t)+8))\)

Note that

\((\frac{d}{dt})(5\ln(\sin(t)+8))=5\frac{d}{dt}(\ln(\sin(t)+8)) [∵(c⋅f) ]\)

\(=5\frac{d}{dt}(\ln u), =5\frac{d}{dt}(\ln (u))(\frac{d}{dt})(\sin(t)+8) =5*(\frac{1}{u})\cos(t) =\frac{5\cos(t)}{\sin(t)+8}/\)

Hence we get

\(\frac{d}{dt}(5\ln(t)+8))=\frac{5\cos(t)}{\sin(t)+8}\)