= lim
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= lim f (x + h) lim
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= f (x) Dx g(x) + g(x) Dx f (x) In the last step, lim f (x + h) = f (x) follows from the fact that f is continuous at x, which is a consequence of Theorem 13.1.
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13.2 Prove Rule 6, the quotient rule:
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If f and g are differentiable at x and if g(x) = 0, then Dx f (x) g(x) = g(x) Dx f (x) f (x) Dx g(x) [g(x)]2
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If g(x) = 0, then 1/g(x) is de ned. Moreover, since g is continuous at x (by Theorem 13.1), g(x + h) = 0 for all suf ciently small values of h. Hence, 1/g(x + h) is de ned for those same values of h. We may then calculate lim
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1 1 g(x+h) g(x)
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= lim = lim
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g(x) g(x + h)
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h 0 hg(x)g(x + h)
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[by algebra:
multiply top and bottom by g(x)g(x + h)] [by algebra] [by Property IV of limits]
1/g(x) h 0 g(x + h)
g(x + h) g(x) h
= lim = = = Having thus proved that
1/g(x) g(x + h) g(x) lim g(x + h) h 0 h h 0
lim [ 1/g(x)]
lim g(x + h)
Dx g(x)
[by Property VI of limits and differentiability of g]
1/g(x) Dx g(x) g(x) 1 Dx g(x) [g(x)]2
[by Property II of limits and continuity of g]
1 g(x)
1 Dx g(x) [g(x)]2 1 1 Dx f (x) Dx g(x) + g(x) [g(x)]2
we may substitute in the product rule (proved in Problem 13.1) to obtain Dx f (x) g(x) = Dx f (x) = = which is the desired quotient rule. 1 g(x) = f (x)
f (x)Dx g(x) g(x)Dx f (x) + [g(x)]2 [g(x)]2 g(x)Dx f (x) f (x)Dx g(x) [g(x)]2
MORE ON THE DERIVATIVE
[CHAP. 13
13.3 Prove Rule 7: Dx (x k ) = kx k 1 for any integer k.
When k is positive, this is just Rule 4 ( 12). When k = 0, Dx (x k ) = Dx (x 0 ) = Dx (1) = 0 = 0 x 1 = kx k 1 Now assume k is negative; k = n, where n is positive. algebra By de nition, 1 x k = x n = n x By (1) of Problem 13.2, Dx (x k ) = Dx 1 xn = 1 xn
Dx x n
But (x n )2 = x 2n and, by Rule 4, Dx (x n ) = nx n 1 . Therefore, 1 Dx (x k ) = 2n nx n 1 = nx (n 1) 2n = nx n 1 = kx k 1 x algebra We have used the law of exponents, xa = x a b xb
13.4 Find the derivative of the function f such that f (x) =
Use the quotient rule and then Rule 5 ( 12), Dx x2 + x 2 x3 + 4 = = = =
x2 + x 2 . x3 + 4
(x 3 + 4)Dx (x 2 + x 2) (x 2 + x 2)Dx (x 3 + 4) (x 3 + 4)2 (x 3 + 4)(2x + 1) (x 2 + x 2)(3x 2 ) (x 3 + 4)2 (2x 4 + x 3 + 8x + 4) (3x 4 + 3x 3 6x 2 ) (x 3 + 4)2 x 4 2x 3 + 6x 2 + 8x + 4 (x 3 + 4)2
13.5 Find the slope-intercept equation of the tangent line to the graph of y = 1/x 3 when x = 1 . 2
The slope of the tangent line is the derivative dy 1 = Dx 3 dx x When x = 1 , 2 dy 3 3 = = 1 = 3(16) = 48 1 4 dx x=1/2 16 2 So, the tangent line has slope-intercept equation y = 48x + b. When x = 1 , the y-coordinate of the point on the graph is 2 1
1 3 2
3 = Dx x 3 = 3x 4 = 4 x
1 = 1 =8
CHAP. 13]
MORE ON THE DERIVATIVE
Substituting 8 for y and 1 for x in y = 48x + b, we have 2 8 = 48 Thus, the equation is y = 48x + 32. 1 +b 2 or 8 = 24 + b or b = 32
