f（x）=（x^2+x+3）^48（x^3+5x-2）^84 f'（x）=？

f（x）=（x^2+x+3）^48（x^3+5x-2）^84 f'（x）=？

48（2x+1）（x^2+x+3）^47（x^3+5x-2）^84+84（3x^2+5）（x^2+x+3）^48（x^3+5x-2）^83
應該就這個了，

Alevel數學題？ 1.P is the point（7.5）and L1 is the line with equation 3x+4y=16 （1）.Find out the equation of the line L2 which passes through P and is perpendicular to L2. （2）.Find the point of intersection of the L1 and L2. （3）.Find the perpendicular distance of P from the line L1.

（1）perpendicular to L1吧？
L2:4x-3y=13
（2）（4,1）
（3）d=（/3*7+4*5-16/）/5=5
（1）問就是求過P且與L1垂直的直線L2
（2）問就是求L1與L2的交點
（3）問就是求P到L1的距離

Simplify the expasion of（1-x）^8+（1+x）^8 including the term in x^2，by putting x=0.01，find the approprizte value of 0.99^8+1.01^8，correct to three decimal places.

（1-x）^8+（1+x）^8 = 2（1+8C2x^2 + 8C4x^4+8C6x^6+x^8）=2（1+ 28x^2+70x^4+28x^6+x^8）= 2+ 56x^2（including the term in x^2）x=0.010.99^8+1.01^8 = 2+56（0.01）^2=2.0056=2.006（3decimal places）

翻譯一道英文出的數學題 Phillip charged $400 worth of goods on his credit card. On his first bill，he was not charged any interest，and he made a payment of $20. He then charged another $18 worth of goods. On his second bill a month later，he was charged 2% interest on his entire unpaid balance. How much interest was Phillip charged on his second bill？ A. $8.76 B. $7.96 C. $7.60 D. $7.24 E. $6.63

（400-20+18）*2%=7.96
The answer is B.
支付了20，又花了18，所以實際花銷是398.

利用夾逼定理證明：若a1，a2，a3，.，am為m個正常數，則 lim（n趨向於∞）n次根號下a1^n+a2^n+.+am^n=A其中A=max{a1，a2，.，am} 利用單調有界數列必存在極限這一收斂準則證明：若x1=根號2，x2=根號下2+根號2，.，xn+1=根號下2+xn（n=1,2，.），則lim（n趨向於∞）xn存在，並求該極限.

第一題：將所有的a1，a2，…，am全部用A代替，這樣把整個式子放大了，結果為
n次根號下（n*A^n）=n次根號下（n）*A，極限為A
然後將該式縮小，a1，a2，…，am中肯定有一個和A相等的，把這一項留下，其餘項删除，這樣就縮小了，結果為：n次根號下（A^n）=A
放大與縮小後的極限都是A，這樣由夾逼準則，本題得證
第二題，首先要證明極限存在，該數列單增是比較顯然的，下麵證明有界，
數學歸納法，x1

1.∫sin^2（2x）就是sin2x的平方 2.∫（sin^x cos^x）

1、變換為對sin（4x）的積分
2、變換為對sin（2x）的積分