The percentage of compression on the lumbar vertebral disks

Last updated: 7/10/2022

The percentage of compression on the lumbar vertebral disks causes strain in an adult human. The strain can be measured by S(x) = 7.2956 In(0.0645012 x^0.95 + 1), where x is the physical load in kilograms (kg). Medical scientists are interested in the rate of change of the strain with respect to the load. This can be found with the derivative S'(x) or ds/dx. The derivative can be found by ... using the derivative rule for Inf(x)) to get 5'(x) = (7.2956) [1/(0.0645012 x 0.95 + 1)] * [(0.0645012) (0.95) x^(-0.05)), which can be simplified to approximately S'(x) = [0.44705] / [(0.0645012 x^0.95 + 1)*(x^ 0.05]] using the Chain Rule to get 5'(x) = 7.2956 In(0.0645012 x 0.95 + 1) ((0.0645012)(0.95) x^(-0.05]], which can be simplified to approximately S'(x) = [0.44705 In(0.0645012 x 0.95 + 1)] / [x^0.05] S'(x) = 7.2956 e^(0.0645012 x^0.95 + 1), where e is the irrational number approximated by 2.71828 using the derivative rule for In(f(x)) to get 5'(x) = (7.2956) [1/(0.0645012 x^0.95 + 1)], which can be simplified to S'(x) = 7.2956/(0.0645012 x^0.95 + 1)