. Differential and integral calculus, an introductory course for colleges and engineering schools. f(x +Ax)= y+Ay is the slope of the tangent to the graph of f(pc) at the point(v,f(a>)). Example 1. To find the derivative of By definition Dx3 = lim AxA From the identitywe have Az=0A£ Af(x)=f(x+Ax)-f(x) AX3 = (X + Az); = 3 x2Ax+ 3 x(Ax)2+ (Ax)3 *. * We use Axn or A(zn) to mean increment of xn, and (Ax)n to mean thenth power of Ax. Similarly, Dxn or D(xn) means the derivative of xn, and(Dx)n means the nth power of Dx. Ax3^AxAX3 lim — = lim (3 x2 + 3 xAx + (Ax)2) = 3 x2. Az=()AX Az
. Differential and integral calculus, an introductory course for colleges and engineering schools. f(x +Ax)= y+Ay is the slope of the tangent to the graph of f(pc) at the point(v,f(a>)). Example 1. To find the derivative of By definition Dx3 = lim AxA From the identitywe have Az=0A£ Af(x)=f(x+Ax)-f(x) AX3 = (X + Az); = 3 x2Ax+ 3 x(Ax)2+ (Ax)3 *. * We use Axn or A(zn) to mean increment of xn, and (Ax)n to mean thenth power of Ax. Similarly, Dxn or D(xn) means the derivative of xn, and(Dx)n means the nth power of Dx. Ax3^AxAX3 lim — = lim (3 x2 + 3 xAx + (Ax)2) = 3 x2. Az=()AX Az=0 Z)x3 = 3 x2. §22Hence and Therefore It will add to the readersunderstanding of this process tonote the geometrical interpreta-tion of the successive steps. Thegraph of x3 is shown in the coordinates of P are x andx3, and those of Q are x + Ax and(x + Ax)3. From the figure it isobvious that Ax3 = (x + Ax)3 — x3, A ., . Ax3 QC ,and that — = — = tan PC THE DERIVATIVE = 3x2 + 3xAx+(Ax)2, 35. (x+Ax) Henceand — = 3x2 + 3xAx + (Ax)2 AX ,. Ax3lim — Ax=0 AX tan a, 3 x2 = lim tan a = tan 0, or DX3 _ 3 X2 = tan 0> Example 2. To find the derivative ofSolution. By definition 7)i = lim-^. £ Ax = 0 AX From the identity A/(x)=/(x+Ax)-/(x) we have *i- 112 xAx + (Ax)2 (x+Ax)2 x2 x2(x+Ax)2 Hence x2 _ 2 x + Ax AX X2(x+Ax)2 and lim Ax=C Ax ,. 2 x+Ax 2x 2ax=o x2(x2+Ax)2 X4 ~ X3 Therefore X2 X3 by e, Art. 20. 36 DIFFERENTIAL CALCULUS §23 Example 3. To determine the derivative of Solution. Let f(x) = Then Af(x) ^ fix + Ax) - f(x) 1 + x2 1 + x2 X + AX Hence l+(x+Ax)2 1 + X2 (1 — x2)Ax— x(Ax)2 (l+X2)[l+(x+Ax)2] Af(x) 1 — X2 — XAX AX ~ (1 + X2) [1 + (x + Ax)2] , y Af(x) _ y 1 — X2— XAX 1 — X2 So AX ~ A% (l + X2)[l + (x+Ax)2]~ (1+X2)2 by e, Art. 20. Therefore finally D —f— = ~ ^ • 1 + x2 (1 + x2)2 From the study of the foregoing examples, it appears that theprocess of finding the derivative of f(x) falls naturally into threesteps: First step. The c
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