To differentiate or calculate the first differential of f(x)='(x) from first principles means to find the first differential of f(x)=f(x) by using the expression Hence d(f(x)) dx f'(x) = - f'(x) = Substitute (x+8x) for x in f(x) = x² to give f'(x) = lim 8x-0 d(f(x)) f(x+8x)-f(x)\ dx Sx f(x+8x)=(x+6x)=x²+2x6x+(8x)² = lim 8x->0 x²+2x8x+(6x)²-x² 8x lim 8x-0 2x8x+(8x)'] Sx au ->> As 8x → 0, [2x + 8x] →→ 2x, so that the differential of x² is 2x. Hence at x =2, the first differential of the gradient of the curve f'(x) = 4. Find lim (2x+8x) dx-0 b. f(x) of f(x) = x³+3x²+5x at x=5.5
To differentiate or calculate the first differential of f(x)='(x) from first principles means to find the first differential of f(x)=f(x) by using the expression Hence d(f(x)) dx f'(x) = - f'(x) = Substitute (x+8x) for x in f(x) = x² to give f'(x) = lim 8x-0 d(f(x)) f(x+8x)-f(x)\ dx Sx f(x+8x)=(x+6x)=x²+2x6x+(8x)² = lim 8x->0 x²+2x8x+(6x)²-x² 8x lim 8x-0 2x8x+(8x)'] Sx au ->> As 8x → 0, [2x + 8x] →→ 2x, so that the differential of x² is 2x. Hence at x =2, the first differential of the gradient of the curve f'(x) = 4. Find lim (2x+8x) dx-0 b. f(x) of f(x) = x³+3x²+5x at x=5.5
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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