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THE CHAIN RULE in Visual Studio .NET
THE CHAIN RULE QR Code Decoder In .NET Framework Using Barcode Control SDK for VS .NET Control to generate, create, read, scan barcode image in VS .NET applications. QRCode Maker In VS .NET Using Barcode encoder for .NET framework Control to generate, create QR Code ISO/IEC18004 image in Visual Studio .NET applications. [CHAP. 15
Denso QR Bar Code Scanner In .NET Using Barcode decoder for Visual Studio .NET Control to read, scan read, scan image in .NET framework applications. Barcode Encoder In VS .NET Using Barcode generation for .NET framework Control to generate, create barcode image in .NET framework applications. (2) Substitute a = (x + h)1/k and b = x 1/k in step (1): 1 (x + h)1/k x 1/k = (k 1)/k + (x + h)(k 2)/k x 1/k + + (x + h)1/k x (k 2)/k + x (k 1)/k h (x + h) (3) Let h 0 in step (2). 15.27 Prove the chain rule (Theorem 15.3): (g f ) (x) = g ( f (x))f (x), where f is differentiable at x and g is differentiable at g(y + t) g(y) g (y) for t = 0. Since t lim G(t) = 0, let G(0) = 0. Then g(y + t) g(y) = t(G(t) + g (y)) holds for all t. When t = K, g(y + K) g(y) = K(G(K) + g (y)) g( f (x + h)) g( f (x)) = K(G(K) + g (y)) So, K = lim K H(x + h) H(x) = (G(K) + g (y)) h h Recognize Barcode In VS .NET Using Barcode reader for .NET framework Control to read, scan read, scan image in .NET applications. QR Code 2d Barcode Generator In C# Using Barcode generation for Visual Studio .NET Control to generate, create QR Code image in .NET framework applications. f (x). [Hint: Let H = g f . Let y = f (x) and K = f (x + h) f (x). Also let G(t) = QR Code Printer In Visual Studio .NET Using Barcode creation for ASP.NET Control to generate, create Denso QR Bar Code image in ASP.NET applications. Printing QRCode In Visual Basic .NET Using Barcode drawer for VS .NET Control to generate, create QRCode image in VS .NET applications. Now lim
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USS Code 39 Creation In None Using Barcode encoder for Online Control to generate, create Code 39 image in Online applications. Drawing Code128 In Java Using Barcode encoder for BIRT Control to generate, create Code128 image in BIRT applications. A function is usually de ned explicitly by means of a formula. EXAMPLES
Matrix 2D Barcode Drawer In C#.NET Using Barcode generator for .NET framework Control to generate, create Matrix Barcode image in VS .NET applications. Reading Bar Code In None Using Barcode scanner for Software Control to read, scan read, scan image in Software applications. (a) f (x) = x 2 x + 2 (b) f (x) = x (c) f (x) = x2 1 1 x if x 1 if x < 1
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(a) The equation y3 x = 0 implicitly determines y as a function of x. In this case, we can solve for y explicitly, y3 = x or y= 3 x (b) The equation y3 + 12y2 + 48y 8x + 64 = 0 is satis ed when y = 2 3 x 4, but it is not easy to nd this solution. In more complicated cases, it will be impossible to nd a formula for y in terms of x. (c) The equation x 2 + y2 = 1 implicitly determines two functions of x, y= 1 x2 and y = 1 x2 The question of how many functions an equation determines and of the properties of these functions is too complex to be considered here. We shall content ourselves with learning a method for nding the derivatives of functions determined implicitly by equations. EXAMPLE Let us nd the derivative of a function y determined by the equation x2 + y2 = 4. Since y is assumed to be some function of x, the two sides of the equation represent the same function of x, and so must have the same derivative, Dx (x 2 + y2 ) = Dx (4) 2x + 2yDx y = 0 2yDx y = 2x Dx y = x 2x = 2y y 133 Copyright 2008, 1997, 1985 by The McGrawHill Companies, Inc. Click here for terms of use.
[by the power chain rule] IMPLICIT DIFFERENTIATION
[CHAP. 16
Thus, Dx y has been found in terms of x and y. Sometimes this is all the information we may need. For example, if we want to know the slope of the tangent line to the graph of x 2 + y2 = 4 at the point ( 3, 1), then this slope is the derivative Dx y = 3 x = = 3 y 1 The process by which Dx y has been found, without rst solving explicitly for y, is called implicit differentiation. Note that the given equation could, in this case, have been solved explicitly for y, y = 4 x2 and from this, using the power chain rule, 1 Dx y = Dx ( (4 x 2 )1/2 ) = (4 x 2 ) 1/2 Dx (4 x 2 ) 2 x x 1 = ( 2x) = = 2 2 2 4 x 4 x 4 x2 Solved Problems
16.1 Consider the curve 3x 2 xy + 4y2 = 141. (a) Find a formula in x and y for the slope of the tangent line at any point (x, y) of the curve. (b) Write the slopeintercept equation of the line tangent to the curve at the point (1, 6). (c) Find the coordinates of all other points on the curve where the slope of the tangent line is the same as the slope of the tangent line at (1, 6). (a) We may assume that y is some function of x such that 3x 2 xy + 4y2 = 141. Hence, Dx (3x 2 xy + 4y2 ) = Dx (141) 6x Dx (xy) + Dx (4y2 ) = 0 6x x dy dy + y 1 + 8y =0 dx dx x dy dy + 8y = y 6x dx dx dy ( x + 8y) = y 6x dx y 6x dy = dx 8y x which is the slope of the tangent line at (x, y). (b) The slope of the tangent line at (1, 6) is obtained by substituting 1 for x and 6 for y in the result of part (a). Thus, the slope is 6 6 0 6 6(1) = = =0 8(6) 1 48 1 47 and the slopeintercept equation is y = b = 6. (c) If (x, y) is a point on the curve where the tangent line has slope 0, then, y 6x =0 8y x or y 6x = 0 or y = 6x CHAP. 16]

