limitation and differentiation

Anonymous

2013-05-02 8:09 am

1. Differentiate with respect with x.

x^ ln (x^2)

2. Evaluate lim ( 1 + x - e^sin x) / x^2 when x tends to 0.

3. Show from the first principle that d/dx (cosx) = - sin x

Hint: cos A - cos B) = -2 sin [(A+B)/2] sin [(A-B)/2]

回答 (1)

50418129

2013-05-02 4:45 pm

✔ 最佳答案

(1)

let y = x^ln(x^2)
ln y = ln (x^2) ln x = 2(ln x)^2
(1/y)(dy/dx) = 4(ln x)/x
dy/dx = 4y(ln x)/x or 4(ln x)[x^ln(x^2) - 1]

(2)

Let y = e^sin x
ln y = sin x
(1/y)(dy/dx) = cos x
dy/dx = ycos x = (e^sinx) cos x

lim(x->0) (1 + x - e^sin x) / x^2
= lim(x->0) [0 + 1 - (e^sinx) cos x] / 2x
= lim(x->0) {- [(e^sinx) cos x](cos x) - (-sin x)(e^sinx)} / 2
= (- 1 x 1 x 1 + 0 x 1) / 2
= -1/2

(3)

d/dx (cosx)
= lim(h->0) [cos(x + h) - cos x] / h
= lim(h->0) - 2 sin[(2x + h)/2] sin(h/2) / h
= lim(h->0) - 2 sin(x + h/2) sin(h/2) / h
= lim(h->0) - sin(x + h/2) sin(h/2) / (h/2)
= - sin(x + 0/2) (1)
= - sin x

參考: knowledge

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