If a ball is thrown vertically upward with a velocity of 72 ft/s, then its height after t seconds is s(t)=64t-16t^2.
1. What is the maximum height reached by the ball? Explain why this location is the maximum using the derivative as part of your answer.
2. find the velocity of the ball right...
Find the first and second derivative--simplify your answer.
y=x tanx
I solved the first derivative.
y'=(x)(sec^2(x)) +(tanx)(1)
y'=xsec^2(x) +tanx
I don't know about the second derivative though.