Use the definition of the derivative to show how the formula

Use the definition of the derivative to show how the formulas in Problems 34-43 are obtained. If f(x) =5x, then f\'(x) = 5. If f(x) = 3x - 2, then f\'(x) = 3. If f(x) = x^2 + 4, then f\'(x) = 2x. If f(x) = 3x^2, then f\'(x) = 6x. If f(x) = x - x^2, then f\'(x) = 1 - 2x. If f(x) = 5x^2 + 1, then f\'(x) = 10x. If f(x) = 2x^2 + x, then f\'(x) = 4x + 1. If f(x) = -2x^3, then f\'(x) = -6x^2. If f(x) = 1 - x^3 then f\'(x) = -3x^2. If f(x) = 1/x, then f\'(x) = -1/x^2.

Solution

34) f(x) = 5x , then f\'(x) = 5

according to definition of derivative , the derivative of f(x) with respect to x is given by

f\'(x) lim h ---> 0 = f(x+h) - f(x) / h

so f(x+h) = 5(x+h)

plugging the values in the formula

f\'(x) = 5x + 5h - 5x / h = 5

hence, f\'(x) = 5

35) f(x) = 3x - 2 , f\'(x) = 3

f(x+h) = 3(x+h) - 2 = 3x + 3h - 2

plugging the values in the formula

f\'(x) = { 3x + 3h - 2 - 3x + 2 } / h = 3

f\'(x) = 3

36) f(x) = x^2 + 4

f(x+h) = (x+h)^2 + 4 = x^2 + h^2 + 2xh + 4

plugging the values in the formula

f\'(x) = { x^2 + h^2 + 2xh + 4 - x^2 - 4 } / h

f\'(x) lim h ---> 0 = 2x

37) f(x) = 3x^2

f(x+h) = 3(x+h)^2 = 3(x^2 + h^2 + 2xh ) = 3x^2 + 3h^2 + 6xh

f\'(x) lim h-->0 = {3x^2 + 3h^2 + 6xh - 3x^2} / h   = 6x

f\'(x) = 6x