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Examville is a global education community where users like you can connect and interact with other students and teachers from around the world. Share, seek, download and discuss everything inside and outside the classroom. All you need is an email address and a password to get started. JOIN US FOR FREE: http://www.examville.com © Examville.com, LLC June 2009 DIFFERENTIATION Let us take the graph of a function f . We know that the point ( x,f(x) ) is a point on the graph. What line, if any, should be the tangent to the graph at this point? We can take another point on ( x+h, f(x+h) ) on the graph and draw a secant line through these two points. Keeping ( x,f(x) ) fixed, we move the other point closer to it (h tends to zero from the left or from the right). The secant line tends to a limiting position. This is “the tangent to the graph at the point ( x,f(x) )”. If this limit exists, that is defined to be the slope of the tangent line at that fixed point ( x,f(x) ) on the graph of y = f (x). DIFFERENTIABILITY & THE DERIVATIVE OF A FUNCTION Differentiation is the process of finding the derivative of a function. A function f is said to be differentiable at x if and only if ffff+fffffffff ` a ` a x fh @f x ffffffffffff ffffffffffff ffffffffffff lim exists. hQ0 h ` a If this limit exists, it is called the derivative of f at x and is denoted by f. x . x + f fff f ` a ` a ffffffffffxff ffffhf@ f fff ffffffffffffff fffffffffffff Therefore, f. x = lim ` a . hQ 0 h ` a If we draw a tangent to the graph at this point, the slope of the tangent is equal to f. x . The derivative of a function y = f(x) with respect to x at a point x = x 0 is given by x 0 + f ff x 0 b c ` a f ffffffhf@ffffff fffffffffffff ffffffff fffff f. x 0 = lim ffffffffffffff provided that this limit exists. ` a hQ0 h This limit is also called the instantaneous rate of change of y with respect to x at x = x 0 . ` a d f ` a dyf ff ff f ff ff f The d/dx notation : If y = f(x), then we can write f. x = fff x = ff dx dx The derivative of the function y =f(x) with respect to x may be indicated by any one of ` a dff ff` a ff ff dyf ff f ff ff ` a dfffff ffx f f ffff fffff the symbols y , , y. , f. x , Dx , Dx f, . We can use any of these, dx dx dx while solving problems. • A function is said to be differentiable at a point x = x 0 if the derivative of the function exists at that point. • Also, if a function is differentiable at every point in an interval, the function is said to be differentiable on that interval. • If we say “a function (of x) is differentiable” without mentioning the interval, we mean that it is differentiable for every value of x in its domain. Example : f x = x 3 @ 3x ` a 1. The function is differentiable on the interval (-1,1) as it is differentiable at every point in that interval. 2. The function f x = |x @ 2| is not differentiable on the interval (1,3) as it is not ` a differentiable at x = 2 which lies in that interval. Example : Find f’(x) for the function f x = x 2 + 4 . ` a Solution : We formthe difference quotient +fffffffffffff4 f @ f2 + f hff+fffffffffff B` a2 C B C ffff+fffffffff fffffffff xfffffffffffxfffff 4ffff fffff ` a ` a x fh @f x ffffffffffff fff ffffffffffff f ffffffffffff fff f f = h h + 2xh + h + 4 @ x 2 @ 4 ff+ h 2 2 2 xfffffffffffffffffffff 2xhfffff fffffffffffffffffffff fffffff ffffffffffffffffffff ffffff fffffffffffffffffffff fffffff = = = 2x + h h h x + h @f x ` a ` a fffffffffffff ffffffffffff ffffffffffff Therefore f. x = lim fffffffffffff lim 2x + h = 2x = ` a ` a hQ0 h hQ0 Example : Find f. @ 3 and f. @ 1 given that f x = x 2 ` a ` a ` a Solution : @ f + h fff @ 3 ` a ` a ffffffffff fffff ffff3 ffff@fffffff fffffffffffffff By definition f. @ 3 = lim ffffffffffffffff ` a hQ0 h f3 + h @ @ 3 ` a2 ` a2 @ffffffffffffff ffffffffffffff ffffffffffffff fffffffffffffff = lim hQ0 h c @ 6h + h ff b 2 9 fffffffff@ff ffffffffffffff ffffffffffffff fffffffffff f f 9 = lim hQ0 h 6h + h 2 @fffffff ffffffff fffffff ffffffff = lim hQ0 h a = lim @ 6 + h ` hQ0 =@6 Similarly @ f + h fff @ 3 ` a ` a ffffffffff fffff ffff1 ffff@fffffff fffffffffffffff f. @ 1 = lim ffffffffffffffff ` a hQ0 h f1 + f @ fff ` a2 ` a2 ffffffffff 1 ff fffffhffff@ffff @fffffffffffff ffffffffffffff = lim hQ0 h = lim @ 2 + h ` a hQ0 =@2 ` a ` a Alternative method : Find f. x from f(x) first, and to put x=-3 and x=-1 in f. x to get ` a ` a f. @ 3 and f. @ 1 as shown below ffff+fffffffff ` a ` a x fh @f x ffffffffffff ffffffffffff ffffffffffff f. x = lim ` a hQ0 h xf+ffffffff ` a2 2 fh @x ffffffffff ffffffffff ffffffffff = lim hQ0 h +fffffff +fffffffffffff b c 2 2 h @ f2 2xhfffffffff xfffffffffffffxff fffffff fff f fffffff f = lim hQ0 h ff+ h 2 2xhfffff fffffff ffffff fffffff = lim hQ0 h = lim 2x + h ` a hQ0 = 2x Therefore f. @ 3 = 2 A @ 3 = @ 6 ` a ` a and f. @ 1 = 2 A @ 1 = @ 2 ` a ` a Example : Find the derivative of f x = x 3 + 5 at the point (2,13). ` a Solution : f x = x3 + 5 ` a x + h @f x ` a ` a fffffffffffff = ffffffffffff ffffffffffff lim fffffffffffff ` a f. x hQ0 h B` C B C hff+fffffffffff @ f3 + f +fffffffffffff5 f a3 5ffff fffff xfffffffffffxfffff = fff f fff f f lim fffffffff hQ0 h + fff + f + f fffffff +fffffff3xhfffhffffffffffff b c 3 2 2 3 hfffffffffffffff x 3 @ f 3xffffffffffffffff@ffffff xfffffffffffffffffff5fffffff5f = fff f fffff lim ffffffff hQ0 h f h + 3xh + h 2 2 3 3xfffffffffffff = fffffffffffff fffffffffffff lim ffffffffffffff hQ0 h = lim 3x 2 + 3xh + h b c 2 hQ0 = 3x 2 Therefore f. 2 = 3.2 = 12 ` a 2 Hence the derivative of f x = x 3 + 5 at the point 2,13 is 12 A ` a b c COMPARISON OF DIFFERENTIABILITY AND CONTINUITY If a function f is differentiable x, it is continuous at x. But if a function is continuous at x, it may or may not be differentiable at x. Example : 1. The function f x = x 3 @ 3x is differentiable on its whole domain (which is R). ` a As differentiability implies continuity, it is continuous at every point. 2. f(x) = |x| is continuous at any real value of x. But as continuity does not imply differentiability, we have to check its differentiability at every at every point. We find that at x=0, it is not differentiable though continuous. 2 ` a xfffff @4 3. The function ffff fffff f x = fffff is not continuous at x = 2 . As continuity is x @2 necessary for differentiability, it is not differentiable at that point. DIFFERENTIATION FORMULAS We can use the limit definition of derivative to find the derivative of any function, but this application may be cumbersome and very long at times. For example, finding the f+fff@ 9 f 2 f 5x fff xfffffffff fffffffff fffffffff derivative of by the limit method will be very long. So, many useful x @x + 2 3 differentiation formulas are first proven using the definition of derivative, and then these are used directly in solving problems. DIFFERENTIATION OF COMBINATION OF FUNCTIONS The functions to be differentiated are usually combination of two or more functions. Here we discuss about the combination of two functions by addition, subtraction, multiplication, division only and also scalar multiples of functions. (Differentiation of composite functions are discussed separately in Chain Rule). ` a ` a ` a In the following formulas u , v and w, or f x , g x , h x are functions of x and are differentiable, and c is a constant number dff ff` ff ff duf ff ff cu = c A fff a Scaler multiple of a function : dx dx dff ff ff ff c A f x = c f. x B ` aC ` a [ dx dff duf dvf ff ff ff ff ff ff f Sum and Difference Rule : ff u F v = ffF ff f ` a dx dx dx dff ff ff ff ` a f x0 F g x = f. x F g. x B ` aC ` a ` a [ dx dff ff ff` ff dvf duf ff ff f uv = u A ff+ v A ff ff f ff a Product Rule : dx dx dx dff ` a ` aC ` a ` a ` a ` a ff ffff f x A g x = g x A f. x + f x A g. x B [ dx we can also extend this to product of three functions as dff ff ff ff` dwf dvf duf ff ff f uvw = uv A fff uw A ff + vw A ff + f ff ff f ff ff a dx dx dx dx du dv dff f ff uf d e ff f ff f A fff@ffffff ff f ff ff vffffffffffff u A fff f fff fffffffffff ffffffffffff f ff ff ff dx dx Quotient Rule : = dx v v2 H ` aI ` a ` a ` a ` a x dff fffff ff fff ff ffff ffJ ffff ffffffff@ f ffffffx f gfffff.ffffffffffg.ffff x A f x fff x A ff f ffffffffffffffffffff ffffffffffffffffffff [ ` aK = B ` aC2 dx g x g x DIFFERENTIATION OF x n AND ` a c CONSTANT dff ff ff ff` a c =0 f x = c, then f. x = 0 ` a ` a that is, if dx dff ff ff ff` xn = n A xn@ 1 a dx using this we can write the derivative of another two common functions dff ff ff ff` a x =1 dx ff 1f dff f 1f f g ff f ff f ff ff = @ ff dx x x2 1f 1f f ff Example : Differentiate y = fx 3 @ fx 2 + x @ 3 6 4 Answer : dyf ff 1f 3 1f 2 ff d f f f g ff ff f f ff ff f f ff = x @ x + x @3 using the sum @ difference rule dx dx 6 4 dff f ff 1f 3 ff 1f 2 dff f dff dff f g f g ff ff f ff f ff f ff` a ff ff ff ff ff` a = x @ x + x @ 3 dx 6 dx 4 dx dx 1fb c 1f ` a ` a ` a ff fff = A 3x 2 @ A 2x + 1 @ 0 6 4 2 xff xf ff f f ff f f = @ +1 2 2 Example : Differentiate y = x 2 @ 2x + 3 x 3 @ x 2 + 2 b cb c Answer : dyf ff b 2 ff d f D E ff ff f ff ff = f x @ 2x + 3 x 3 @ x 2 + 2 cb c using the product rule dx dx c d b c d b ff ff ff ff ff ff = x 3 @ x 2 + 2 A ff x 2 @ 2x + 3 + x 2 @ 2x + 3 A ff x 3 @ x 2 + 2 b c b c dx dx = x @ x + 2 A 2x @ 2 + x @ 2x + 3 A 3x @ 2x b c` a b 2 cb c 3 2 2 = 2 x @ 1 x 3 @ x 2 + 2 + x 3x @ 2 x 2 @ 2x + 3 ` ab c ` ab c f+ff 3 xfffff f1 fffff Example : Differentiate y = fffff 3 x @1 Answer : ff ff f+ff 3 dyf ff xfffff ff d f fffff 1 f g f ff ff fffff = f 3f using the quotient rule dx dx x @ 1 d d 1 A fffx 3 + 1 @ x 3 + 1 A fffx 3 @ 1 b c b c b c b c x f@ffffffffffffffffffffffffffffff 3 f ff ff f ff f ffffffffffffffffffffffffffffffffff fff fffffffffffffffffffffffffffff ffffffffffffffffffffffffffffffffff ff ff f f dx dx = b c2 x3@ 1