Derivative of Logarithm Function If y = logex then = dy dx 1 x Proof Given: y = logex then x = ey dx dy = ey Exercise 4.6 (page 146) dy dx = 1 ey = 1 x

Function of a Function Rule If y = logef(x) then = f’(x). = dy dx 1 f(x) f’(x) Proof Let u = f(x) then y = logeu du dx Also = f’(x) dy du = 1 u dy dx du = x Exercise 4.6 (page 146) = . f’(x) 1 u = .f’(x) 1 f(x) f’(x) f(x) =

Function of a Function Rule – Example 1/3 f’(x) f(x) = dy dx Differentiate loge(2x2 - x + 1) y = loge(2x2 - x + 1) Exercise 4.6 (page 146) 4x – 1 dy dx = 2x2 - x + 1

Function of a Function Rule – Example 2/3 f’(x) f(x) = dy dx x + 1 x - 1 Differentiate loge = y = loge(x + 1) - loge(x - 1) 1 1 (x - 1).1-(x + 1).1 (x + 1)(x - 1) = dy dx = x + 1 x - 1 Exercise 4.6 (page 146) x – 1 - x - 1 (x + 1)(x - 1) = - 2 x2 - 1 =

Function of a Function Rule – Example 3/3 f’(x) f(x) = dy dx Find the gradient of the normal to loge(x2 - x + 1) @ x = 1 g(x) = loge(x2 - x + 1) Gradient of Normal m2 = -1 m1 2x – 1 g’(x) = Exercise 4.6 (page 146) x2 - x + 1 2 – 1 g’(1) = 1 - 1 + 1 = -1 1 m1 = 1 m2 = -1