Lecture 17 MATH10070 - Introduction to Calculus maths.ucd.ie/modules/math10070 Kevin Hutchinson 28th October 2010
Z Chain Rule (I): If y = f (u) and u = g(x) dy dx = dy du du dx Z Chain rule (II): d dx (f g(x)) = f (g(x)) g (x).
Example Differentiate f (x) = 1 3 x 7 + 7 We have f (x) = (x 7 + 7) 1/3. Thus f (x) = (1/3) (x 7 + 7) 4/3 7x 6. 7x 6 = 3(x 7 + 7) 4/3.
The Generalised Power Rule This is a special - but very often used - case of the Chain Rule. Z Generalised Power Rule If u = f (x) is a function of x, then or d dx (ur ) = ru r 1 du dx d dx f (x)r = rf (x) r 1 f (x)
Example The side, S of a cubical container is 10cm and is increasing at a rate of 0.5cm per second. At what rate is the volume of the container increasing? Solution: Volume = V = S 3. Therefore, the rate of increase of volume is dv dt = d dt S3 = 3S 2 ds dt = 3 10 2 0.5 = 150cm 3 per second.
Higher Derivatives Given y = f (x), differentiation yields a new function dy dx = f (x) This function, in turn, can be differentiated to get: d dx ( ) dy dx the second derivative of f (x). = d 2 y dx 2 = f (x)
Continuing the process we get d 3 y dx 3 = f (x) = f (3) (x) d 4 y dx 4 = f (4) (x) =... etc. the third, fourth and higher derivatives of y = f (x).
Example If s = s(t) is the position of a body at time t, then ds dt = s (t) is the velocity ( = rate of change of position) at time t. Thus d 2 s dt 2 = s (t) is the rate of change of velocity at time t; i.e., it is the acceleration at time t.
Example If s(t) = 5t 3 2t 2 + 3t, find the acceleration at time t. Solution: s (t) = 15t 2 4t + 3. So s (t) = 30t 4 is the acceleration at time t.
Example Find d 2 y dx 2 where y = 6x 4 + 11x 3 + 7x 2 + 3 dy dx = 24x 3 + 33x 2 + 14x d 2 y dx 2 = 72x 2 + 66x + 14
Example Find d 2 y dx 2 where y = 3 x. y = x 1/3. So Hence dy dx = 1 3 x 2/3 d 2 y dx 2 = 1 3 2 3 x 5/3 = 2 9 3 x 5.
Example Consider the function y = 1/x. Using the power and constant rules each time, we obtain: dy dx = 1 2 = x x 2 d 2 y dx 2 = ( 2x 3 ) = 2 x 3 d 3 y dx 3 = 2 ( 3x 4 ) = 6 x 4 d 4 y dx 4 = 6 4x 5 = 24 x 5 =... etc.
Spot the general pattern The signs alternate: the odd higher derivatives have a minus. The power of x in the denominator increases by 1 each time. In the nth derivative we get x n+1 in the denominator The nth numerator is obtained by multiplying the previous numerator by n Thus: d n y dx n = n! ( 1)n x n+1 where n! = 1 2 (n 1) n.
Curve sketching A graph gives a much better idea of how a function behaves than its algebraic formula We are going to look at how to produce a sketch of a function s graph from its formula We already know how to sketch straight lines and quadratic functions (see Lecture 4 and Lecture 5) We will be sketching other differentiable functions... but we start with some elementary facts
Positive and Negative slope If the slope of a line is a positive number, the line makes an angle less than 90 with the x-axis (leans forward, /) If the slope is negative then the line makes an angle greater than 90 (leans backwards, \) Z A positive slope to a line means the line is rising (increasing) as we go from left to right, negative slope means the line is decreasing as we go from left to right.
Recall that the derivative of a function f at a point x is just the slope of the tangent line thus knowing whether the derivative is positive or negative tells us something about the graph of f a positive derivative seems to indicate the values of the function are increasing, and a negative derivative that the function is decreasing
Increase, decrease and critical points Definition Let f : R R be a differentiable function and let a R. 1. We say that f is INCREASING at a if f (a) > 0, and that f is DECREASING at a if f (a) < 0. 2. If I is an interval such that f is increasing at every point of I, then I is called an INTERVAL OF INCREASE for f. If f is decreasing at every point of I then I is called an INTERVAL OF DECREASE for f. 3. If f (a) = 0 then a is called a CRITICAL point of f.
Increasing f is increasing at a f (a) > 0 the tangent line to the graph of f has positive slope the tangent line leans forwards (/) Decreasing f is decreasing at a f (a) < 0 the tangent line to the graph of f has negative slope the tangent line leans backwards (\) Critical Point f (a) = 0 the tangent line to the graph of f at a is horizontal a is a root of f (x)
Illustration