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(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
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f (x). [Hint: Let H = g f . Let y = f (x) and K = f (x + h) f (x). Also let G(t) =
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f (x + h) f (x) = f (x). Since lim K = 0, lim G(K) = 0. h h 0 h 0 Hence, H (x) = g (y)f (x) = g ( f (x))f (x).]
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Implicit Differentiation
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A function is usually de ned explicitly by means of a formula. EXAMPLES
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(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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However, sometimes the value y = f (x) is not given by such a direct formula. EXAMPLES
(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
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[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 slope-intercept 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 slope-intercept 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]
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