consider the function r(t)= t^t and s(t)=arcos(t), for which you are given the facts that r'(t)=t^t(ln(t)+1) and s'(t)=-1/square root 1-t^2. Do not be connected with where these derivative formulas come from. we restrict our interest in both functions to the domain 0

Question

consider the function r(t)= t^t and s(t)=arcos(t), for which you are given the facts that r'(t)=t^t(ln(t)+1) and s'(t)=-1/square root 1-t^2. Do not be connected with where these derivative formulas come from. we restrict our interest in both functions to the domain 0<t<1. 

1)let w(t)=3t^t-2 arccos (t). Determine w' (t). 
2) find an equation for the tangent line to y=w(t) at the point (1/2 , w(1/2)). 

 3) let u(t)=t^t + arccos(t). is u increasing or decreasing the the instant t=1/2? why?

zepdrix · Answer

$$\Large\bf\sf w(t)=3t^{t-2}\arccos t$$Is this what w(t) looks like?

anonymous · Answer

I don't know how to solve it. I am not sure, but I put it in the derivative calculator and this is what it gave me
3t^t ln(t)+3t^t+2 arc sin(t)