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What is the derivative of h(t)=8sin(piet)+13t^1.3
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is it\[h(t)=8 \pi \cos \pi t+13t ^{1.3}\]
if yes then i can try to solve it
is the first post that you poseted your answer to the problem or are you asking ifthat is the problem
\[\frac{ dh }{dt }=8 \pi \cos \pi t+13 \times 1.3t ^{0.3}\]
ok
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did u catch my solution
is it the last post that you pasted
h(t)=8sin(pie t)+13 t^1.3 d/dt {h(t)}=8piecos(piet)+13(1.3)t^(1.3-1) h'(t)=8pie cos(pie t)+16.9t^0.3
do u need any more explanation @surjit
now how do I find h'(3)
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