which can also be written as: 21 Basic differentiation - Chain rule: (f g) (b) = f (g(b)) g (b) dg dx for f(y) and g(x), = df dy

Transcription

1 2 Basic differentiation - Chain rule: Chain rule: Suppose f and g are differentiable functions so that we can form their composition f g. The derivative of f g at input b can be computed in terms of the derivative of g at input b, and the derivative of f at input g(b). We have which can also be written as: (f g) (b) = f (g(b)) g (b) for f(y) and g(x), Intuition for why chain rule is true. d(f g) dx x=b = df dy y=g(b) dg dx x=b A change of x in the input (from b to b + x) will result in a change in the output of g (from g(b) to g(b)+g (b) x) of g (b) x. In turn a change of g = g (b) x in the input of g(b) into the function f will result in a change of the output of f (g(b)) g = f (g(b))g (b) x.

2 So, the change in the composite f g that results from a x change in the input x is (approximately) f (g(b)) g (b) x. (f g) lim x 0 x f (g(b)) g (b) x = lim x 0 x = f (g(b)) g (b) Examples. We find the derivative of sin(x 2 ) using the change rule. We have sin(x 2 ) = f(g(x)) where f(y) = sin(y) and g(x) = x 2. Then f (y) = cos(y), and g (x) = 2x, so ( sin(x 2 ) ) = cos(x 2 ) 2x. Wefindthederivativeof(sin(x)) 2 usingthechangerule. Wehave(sin(x)) 2 = f(g(x))where f(y) = y 2 and g(x) = sin(x). Then f (y) = 2y, and g (x) = cos(x), so ( (sin(x)) 2 ) = 2sin(x) cos(x).

3 22 Basic differentiation - Inverse function: Suppose f is function with an interval domain D and which is oneto-one and differentiable on the domain. The one-to-one property means there will be an inverse function g. The functions f and g undo each other. b = f(a) g(b) = a point (a,b) is on point (b,a) is on the graph of f the graph of g tangent slope at tangent slope at (a,b) is m (b,a) is m

4 4 (a,b) m (b,a) m -2-4 Derivative of inverse: Suppose f is is one-to-one and differentiable and g is the inverse function. Suppose b = f(a) (so g(b) = a), and f (a) 0. The g is differentiable at b, and g (f(a)) = f (a).

5 Examples. The arcsine function x = (arcsin(y)) is the inverse of the sine function y = sin(x). The derivative of arcsin is: (arcsin(y)) y=sin(x) = (arcsin(y)) = = (sin(x)). y 2 cos(x) = (sin(x)) 2 = y 2, so g ' (b,/m).5 g (inverse) (b,a) (a,b) f 0.5 (a,m) f '

6 The arctangent function x = (arctan(y)) is the inverse of the tangent function y = tan(x). The derivative of arctan is: (arctan(y)) = y=tan(x) = (arctan(y)) = (tan(x)) = ( ) = (cos(x)) 2 ( (cos(x))2 +(sin(x)) 2 ) (cos(x)) 2 ( + (tan(x)) 2 ) = ( + y 2 ), so ( + y 2 ).

7 23 Introductory application of derivatives Examples. A rectangular piece of cardboard is 2 cm 30 cm (approximately A4 size). Squares of size x are cut from the 4 corners and the cardboard is then folded to make a box. The volume V is a function of the side x: V = x(2 2x)(30 2x) = 2x(35 5x + 2x 2 ) = 2(2x 3 5x x) The domain of x is 0 x 2 2 = 0.5 The derivative dv dx is: dv dx = 2 (x2 02x + 35) The quadratic x 2 02x + 35 has roots at 02± = 5±3 79. The derivative is zero at the roots = , and = The root = is not in the domain. The root = will provide a peak in the volume function V. The maximum volume is V x= = =

8 The end of a (east/west) hallway 4 feet wide meets a (north/south) hallway 3 feet wide. See figure. North Wall West Wall θ 4 ft 3ft For each angle θ in the interval (0, π 2 ), determine the longest ladder l(θ) that can fit between the North Wall and West Wall. By geometry l(θ) = l EW (θ) + l NS (θ), where sin(θ) = 4 l EW (θ) and cos(θ) = 3 l NS (θ). So, 4 l(θ) = sin(θ) + 3 with domain the interval (0, π cos(θ) 2 ). Determine the longest ladder that can be turned around the corner. We note that if l(θ a ) > l(θ b ), then a ladder of length l(θ a ) will fit at angle θ a, but NOT at angle θ b. A ladder which can be turned around the corner must be smaller that ALL

9 the lengths l(θ). Therefore, the largest ladder that can be turned around the corner is the smallest value of the function l(θ). We use the derivative to find the smallest value. l(θ) = 4 sin(θ) + 3 l (θ) = 4 ( sin(θ) cos(θ) ) + 3 ( cos(θ) = 4 ( cos(θ) (sin(θ)) 2 ) + 3 ( sin(θ) (cos(θ)) 2 ) = 4(cos(θ))3 + 3(sin(θ)) 3 (sin(θ)) 2 (cos(θ)) 2 ) = 4 ( (sin(θ)) (sin(θ)) 2 ) + 3 ( (cos(θ)) (cos(θ)) 2 ) = 3 cos(θ) ( (tan(θ)) 3 ) (sin(θ)) 2 The part 3 cos(θ) (sin(θ)) is positive. So, the sign of 2 (l(θ)) is determined by the part (tan(θ)) 3. On the domain (0, π 2 ), the function tan increases. Therefore, the function (tan(θ)) 3 is increasing it starts off negative, reaches zero and then is postive. So the tangent slopes are negative, then zero at some point, and then positive. The place θ 0 where the derivative equals zero is: 0 = (tan(θ 0 )) 3, so tan(θ 0 ) = ( 4 3 ) 3, so θ 0 = arctan( 4 3 ) 3 = While the tangent slopes are negative, the function is decreasing. When the tangent slopes are positive the function is increasing.

10 The smallest value of the function l, is l(arctan( 4 3 ) 3) = (feet). This smallest value of l is the longest ladder that can be turned around the corner.