1st one:differentiatey=(x^3/2)+(48/x)

2nd one: The fixed point A has coordinates (8, -6,5) and the variable point P has coordinates (t,t2t).

a. Show that AP^2=6t^2-24t+125

b. Hence find the value of t for which the distance AP is least.

c. Determine this least distance.

thank u