Can some one please explain these both! Thanks so much The population of bacteria (in millions) in a certain culture x hours after an experimental35xnutrient is introduced into the culture is P(x) =Use the differential to approximate the changes in population for the following changes in x.5+xa. 2 to 2.5b. 5 to 5.25a. Use the differential to approximate the change in population for x=2 to 2.5.Between 2 and 2.5 hours, the population of bacteria changes by million.(Round to three decimal places as needed.)b. Use the differential to approximate the change in population for x = 5 to 5.25.Between 5 and 5.25 hours, the population of bacteria changes bymillion.(Round to three decimal places as needed.)Enter your answer in each of the answer boxes. dyfor 3xe y = 6- In 16+4ln x, with the conditionsdx= 6, x=2, y=0.Assume x and y are functions of t. EvaluatedtdtdyDdt(Type an exact answer in simplified form.)

Given function is: Px=35x5+x2 The derivation for this above function is: P'x=d35x5+x2dxP'x=35dx5+x2dxP'x=35dxdx·5+x2-x·d5+x2dx5+x22P'x=35x2-2x2+55+x22P'x=-35x2-55+x22A). Change in population from x= 2 to 2.5 dx=∆x=2.5-2dx=∆x=0.5 P'x=dPdx=-35x2-55+x22dP=-35x2-55+x22dx Put the value x=2 and ∆x=0.5 in above equation: dP=-3522-55+2220.5dP=-354-55+420.5dP=3581×0.5dP=0.216 B). Change in population from x= 5 to 5.25: dx=∆x=5.25-5dx=∆x=0.25 P'x=dPdx=-35x2-55+x22dP=-35x2-55+x22dx Put the value x=5 and ∆x=0.25 in above equation: dP=-3552-55+5220.25dP=-3525-55+520.25dP=700100×0.25dP=1.75 C). Given function is: 3xey=6-ln16+4lnx Derivative of this function is: 3xey=6-ln16+4lnxd3xeydt=d6-ln16+4lnxdt3eydxdt+3xeydydt=0-0+41xdxdt ........1 Given conditions are: dxdt=6 , x=2, y=0 Put these in equation (1). 3eydxdt+3xeydydt=0-0+4×1x×dxdt3e06+32e0dydt=0-0+4×12×618+6dydt=12dydt=12-186dydt=-1