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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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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