If f(x) = 9x^10 tan^−1 x, find f '(x).

f(x) = 9x^10 tan^−1 x is a product:  f(x) = (9) * (x^10) * (arctan x).

Use the product rule:  (d/dx) (u*v) = u*dv/dx + v*du/dx and the inverse tangent rule:

(d/dx)(arctan x) =  1 / (1 + x^2).

Then the desired derivative is:

                            1

f '(x) = 9 [x^10* + arctan x*10x^9 ]

                         1+x^2

Note that x^9 can be factored out:

                            x

f '(x) = 9*x^9 [ + 10arctan x ]

                        1+x^2

1 number

if we use a graphing calculator, there is only one such number that satisfies the conclusion of the mean value theorem, that is, a number such that

where the right-hand side is the average rate of change of f on the interval [0.5,0.75], which is the slope of the secant line.

in the attached graph, the black line is the secant line which has its slope as the average rate of change of f on [0.5,0.75], while the orange line is the only tangent line to the graph of f in the interval that satisfies the mvt conclusion.