Solved Examples: Derivatives by Limits
(1.) The tangent line to the graph of f(x) at the point P(0.25, 84) is shown
below.
(a.) What does this tell you about f at the point​ P?
(b.) What is the derivative of f at x = 0.25?
$ (a.) \\[3ex] \text{For point, } P(0.25, 84) \\[3ex] x = 0.25 \\[3ex] y = 84 \\[3ex] y = f(x) \\[3ex] f(x) = y \\[3ex] f(0.25) = 84 \\[3ex] $ The derivative of f at x is the slope of the tangent line at point P
$ (b.) \\[3ex] \text{Point 1 = Point }K = (0, 52.5) \\[3ex] x_1 = 0 \\[3ex] y_1 = 52.5 \\[3ex] \text{Point 2 = Point }P = (0.25, 84) \\[3ex] x_2 = 0.25 \\[3ex] y_2 = 84 \\[3ex] \text{slope } = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] = \dfrac{84 - 52.5}{0.25 - 0} \\[5ex] = \dfrac{31.5}{0.25} \\[5ex] = 126 \\[3ex] \therefore f'(0.25) = 126 $
(a.) What does this tell you about f at the point​ P?
(b.) What is the derivative of f at x = 0.25?
$ (a.) \\[3ex] \text{For point, } P(0.25, 84) \\[3ex] x = 0.25 \\[3ex] y = 84 \\[3ex] y = f(x) \\[3ex] f(x) = y \\[3ex] f(0.25) = 84 \\[3ex] $ The derivative of f at x is the slope of the tangent line at point P
$ (b.) \\[3ex] \text{Point 1 = Point }K = (0, 52.5) \\[3ex] x_1 = 0 \\[3ex] y_1 = 52.5 \\[3ex] \text{Point 2 = Point }P = (0.25, 84) \\[3ex] x_2 = 0.25 \\[3ex] y_2 = 84 \\[3ex] \text{slope } = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] = \dfrac{84 - 52.5}{0.25 - 0} \\[5ex] = \dfrac{31.5}{0.25} \\[5ex] = 126 \\[3ex] \therefore f'(0.25) = 126 $