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الأسئلة

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الوقت

النتيجة

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إذا كان $\sqrt{x}\cdot\tan x$ $u\left(x\right)=$ ، أوجد $u'\left(x\right)$ .

$u'\left(x\right)=\frac{1}{2\sqrt{x}}\cdot\tan x+\sqrt{x}\cdot\sec^2x$

$u'\left(x\right)=12\sqrt{x}\cdot\tan x+\sqrt{x}\cdot\sec^2x$

$u'\left(x\right)=\frac{1}{2\sqrt{x}}\cdot\tan x-\sqrt{x}\cdot\sec^2$

افترض أن $x\cdot\cos x\cdot\ln x$ $w\left(x\right)=$ . أوجد $w'\left(x\right)$ .

$w'\left(x\right)=\cos x\cdot\ln x-x\cdot\sin x\cdot\ln x+x\cdot\cos x\cdot\frac{1}{x}$

$w'\left(x\right)=\cos x\cdot\ln x-x\cdot\sin x\cdot\ln x-x\cdot\cos x\cdot\frac{1}{x}$

$w'\left(x\right)=\cos x\cdot\ln x+x\cdot\sin x\cdot\ln x+x\cdot\cos x\cdot\frac{1}{x}$

إذا كان $\sin x-e^x$ $v\left(x\right)=$ ، فاحسب $v'\left(x\right)$ .

$v'\left(x\right)=\cos x\cdot e^x+\sin x\cdot e^{ }$

$v'\left(x\right)=\cos x\cdot e^x-\sin x\cdot e^x$

$v'\left(x\right)=\cos x\cdot e^x+\sin x\cdot e^x$

افترض أن $x^3\cdot\ln x$ $m\left(x\right)=$ ، احسب m'(x).

$m'\left(x\right)=3x^2.\ln x-x^2.\frac{1}{x}$

$m'\left(x\right)=3x^2.\ln x+x^2.\frac{1}{x}$

$m'\left(x\right)=3x^2.\ln x+\frac{1}{x}$

إذا كانت p(x)=√ ⋅××جوس س، فأوجد p'(x).

$p'\left(x\right)=\frac{1}{2\sqrt{x}}\cdot\cos x+\sin x\cdot\sqrt{x}$

$p'\left(x\right)=\frac{1}{2\sqrt{x}}\cdot\cos x-\sin x$

$p'\left(x\right)=\frac{1}{2\sqrt{x}}\cdot\cos x-\sin x\cdot\sqrt{x}$

أوجد مشتقة $\ln x\cdot e^{-x}$ $t\left(x\right)=$ .

$t'\left(x\right)=\frac{1}{x}\cdot e^{-x}-\ln x\cdot e^{-x}$

$t'\left(x\right)=\frac{1}{x}\cdot e^{-x}+\ln x\cdot e^{-x}$

$t'\left(x\right)=\frac{1}{x}\cdot e^{-x}-\ln x\cdot e^x$

إذا كان $\cos x\cdot\ln x$ $r\left(x\right)=$ ، فأوجد $r'\left(x\right)$ .

$r'\left(x\right)=\sin x\cdot\ln x-\frac{\cos x}{x}$

$r'\left(x\right)=\sin x\cdot\ln x+\frac{\cos x}{x}$

$r'\left(x\right)=\sin x\cdot\ln x+\cos x$

احسب مشتقة $x\cdot e^x\cdot\sin x$ $s\left(x\right)=$ .

$s'\left(x\right)=e^x\cdot\sin x-x\cdot e^x\cdot\cos x-x\cdot e^x\sin x$

$s'\left(x\right)=e^x\cdot\sin x+x\cdot e^x\cdot\cos-x\cdot e^x\sin x$

$s'\left(x\right)=e^x\cdot\sin x+x\cdot e^x\cdot\cos x+x\cdot e^x\sin x$

افترض أن $\frac{1}{x}\cdot e^x$ $q\left(x\right)=$ ، أوجد $q'\left(x\right)$ .

$q'\left(x\right)=-\frac{1}{x^2}\cdot e^x+\frac{1}{x}\cdot e^x$

$q'\left(x\right)=-\frac{1}{x^2}\cdot e^x-\frac{1}{x}\cdot e^x$

$q'\left(x\right)=-\frac{1}{x^2}\cdot e^x+\frac{1}{x}\cdot e^{ }$

أوجد مشتقة الدالة $e^x\cdot\sin x$ $h\left(x\right)=$ .

$h'\left(x\right)=e^x\cdot\sin x+e^x\cdot\cos x$

$h'\left(x\right)=e^x\cdot\sin x-e^x\cdot\cos x$

$h'\left(x\right)=e^x\cdot\sin x+\cos x$

الأسئلة

الوقت

النتيجة

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إذا كان ﻿x⋅tan⁡x\sqrt{x}\cdot\tan xx​⋅tanx﻿ ﻿u(x)=u\left(x\right)=u(x)=﻿ ، أوجد ﻿u′(x)u'\left(x\right)u′(x)﻿ .

﻿u′(x)=12x⋅tan⁡x+x⋅sec⁡2xu'\left(x\right)=\frac{1}{2\sqrt{x}}\cdot\tan x+\sqrt{x}\cdot\sec^2xu′(x)=2x​1​⋅tanx+x​⋅sec2x﻿

﻿u′(x)=12x⋅tan⁡x+x⋅sec⁡2xu'\left(x\right)=12\sqrt{x}\cdot\tan x+\sqrt{x}\cdot\sec^2xu′(x)=12x​⋅tanx+x​⋅sec2x﻿

﻿u′(x)=12x⋅tan⁡x−x⋅sec⁡2u'\left(x\right)=\frac{1}{2\sqrt{x}}\cdot\tan x-\sqrt{x}\cdot\sec^2u′(x)=2x​1​⋅tanx−x​⋅sec2﻿

افترض أن ﻿x⋅cos⁡x⋅ln⁡xx\cdot\cos x\cdot\ln xx⋅cosx⋅lnx﻿ ﻿w(x)=w\left(x\right)=w(x)=﻿ . أوجد ﻿w′(x)w'\left(x\right)w′(x)﻿ .

﻿w′(x)=cos⁡x⋅ln⁡x−x⋅sin⁡x⋅ln⁡x+x⋅cos⁡x⋅1xw'\left(x\right)=\cos x\cdot\ln x-x\cdot\sin x\cdot\ln x+x\cdot\cos x\cdot\frac{1}{x}w′(x)=cosx⋅lnx−x⋅sinx⋅lnx+x⋅cosx⋅x1​﻿

﻿w′(x)=cos⁡x⋅ln⁡x−x⋅sin⁡x⋅ln⁡x−x⋅cos⁡x⋅1xw'\left(x\right)=\cos x\cdot\ln x-x\cdot\sin x\cdot\ln x-x\cdot\cos x\cdot\frac{1}{x}w′(x)=cosx⋅lnx−x⋅sinx⋅lnx−x⋅cosx⋅x1​﻿

﻿w′(x)=cos⁡x⋅ln⁡x+x⋅sin⁡x⋅ln⁡x+x⋅cos⁡x⋅1xw'\left(x\right)=\cos x\cdot\ln x+x\cdot\sin x\cdot\ln x+x\cdot\cos x\cdot\frac{1}{x}w′(x)=cosx⋅lnx+x⋅sinx⋅lnx+x⋅cosx⋅x1​﻿

إذا كان ﻿sin⁡x−ex\sin x-e^xsinx−ex﻿ ﻿v(x)=v\left(x\right)=v(x)=﻿ ، فاحسب ﻿v′(x)v'\left(x\right)v′(x)﻿ .

﻿v′(x)=cos⁡x⋅ex+sin⁡x⋅ev'\left(x\right)=\cos x\cdot e^x+\sin x\cdot e^{ }v′(x)=cosx⋅ex+sinx⋅e﻿

﻿v′(x)=cos⁡x⋅ex−sin⁡x⋅exv'\left(x\right)=\cos x\cdot e^x-\sin x\cdot e^xv′(x)=cosx⋅ex−sinx⋅ex﻿

﻿v′(x)=cos⁡x⋅ex+sin⁡x⋅exv'\left(x\right)=\cos x\cdot e^x+\sin x\cdot e^xv′(x)=cosx⋅ex+sinx⋅ex﻿

افترض أن ﻿x3⋅ln⁡xx^3\cdot\ln xx3⋅lnx﻿ ﻿m(x)=m\left(x\right)=m(x)=﻿ ، احسب m'(x).

﻿m′(x)=3x2.ln⁡x−x2.1xm'\left(x\right)=3x^2.\ln x-x^2.\frac{1}{x}m′(x)=3x2.lnx−x2.x1​﻿

﻿m′(x)=3x2.ln⁡x+x2.1xm'\left(x\right)=3x^2.\ln x+x^2.\frac{1}{x}m′(x)=3x2.lnx+x2.x1​﻿

﻿m′(x)=3x2.ln⁡x+1xm'\left(x\right)=3x^2.\ln x+\frac{1}{x}m′(x)=3x2.lnx+x1​﻿

إذا كانت p(x)=√ ⋅××جوس س، فأوجد p'(x).

﻿p′(x)=12x⋅cos⁡x+sin⁡x⋅xp'\left(x\right)=\frac{1}{2\sqrt{x}}\cdot\cos x+\sin x\cdot\sqrt{x}p′(x)=2x​1​⋅cosx+sinx⋅x​﻿

﻿p′(x)=12x⋅cos⁡x−sin⁡xp'\left(x\right)=\frac{1}{2\sqrt{x}}\cdot\cos x-\sin xp′(x)=2x​1​⋅cosx−sinx﻿

﻿p′(x)=12x⋅cos⁡x−sin⁡x⋅xp'\left(x\right)=\frac{1}{2\sqrt{x}}\cdot\cos x-\sin x\cdot\sqrt{x}p′(x)=2x​1​⋅cosx−sinx⋅x​﻿

أوجد مشتقة ﻿ln⁡x⋅e−x\ln x\cdot e^{-x}lnx⋅e−x﻿ ﻿t(x)=t\left(x\right)=t(x)=﻿ .

﻿t′(x)=1x⋅e−x−ln⁡x⋅e−xt'\left(x\right)=\frac{1}{x}\cdot e^{-x}-\ln x\cdot e^{-x}t′(x)=x1​⋅e−x−lnx⋅e−x﻿

﻿t′(x)=1x⋅e−x+ln⁡x⋅e−xt'\left(x\right)=\frac{1}{x}\cdot e^{-x}+\ln x\cdot e^{-x}t′(x)=x1​⋅e−x+lnx⋅e−x﻿

﻿t′(x)=1x⋅e−x−ln⁡x⋅ext'\left(x\right)=\frac{1}{x}\cdot e^{-x}-\ln x\cdot e^xt′(x)=x1​⋅e−x−lnx⋅ex﻿

إذا كان ﻿cos⁡x⋅ln⁡x\cos x\cdot\ln xcosx⋅lnx﻿ ﻿r(x)=r\left(x\right)=r(x)=﻿ ، فأوجد ﻿r′(x)r'\left(x\right)r′(x)﻿ .

﻿r′(x)=sin⁡x⋅ln⁡x−cos⁡xxr'\left(x\right)=\sin x\cdot\ln x-\frac{\cos x}{x}r′(x)=sinx⋅lnx−xcosx​﻿

﻿r′(x)=sin⁡x⋅ln⁡x+cos⁡xxr'\left(x\right)=\sin x\cdot\ln x+\frac{\cos x}{x}r′(x)=sinx⋅lnx+xcosx​﻿

﻿r′(x)=sin⁡x⋅ln⁡x+cos⁡xr'\left(x\right)=\sin x\cdot\ln x+\cos xr′(x)=sinx⋅lnx+cosx﻿

احسب مشتقة ﻿x⋅ex⋅sin⁡xx\cdot e^x\cdot\sin xx⋅ex⋅sinx﻿ ﻿s(x)=s\left(x\right)=s(x)=﻿ .

﻿s′(x)=ex⋅sin⁡x−x⋅ex⋅cos⁡x−x⋅exsin⁡xs'\left(x\right)=e^x\cdot\sin x-x\cdot e^x\cdot\cos x-x\cdot e^x\sin xs′(x)=ex⋅sinx−x⋅ex⋅cosx−x⋅exsinx﻿

﻿s′(x)=ex⋅sin⁡x+x⋅ex⋅cos⁡−x⋅exsin⁡xs'\left(x\right)=e^x\cdot\sin x+x\cdot e^x\cdot\cos-x\cdot e^x\sin xs′(x)=ex⋅sinx+x⋅ex⋅cos−x⋅exsinx﻿

﻿s′(x)=ex⋅sin⁡x+x⋅ex⋅cos⁡x+x⋅exsin⁡xs'\left(x\right)=e^x\cdot\sin x+x\cdot e^x\cdot\cos x+x\cdot e^x\sin xs′(x)=ex⋅sinx+x⋅ex⋅cosx+x⋅exsinx﻿

افترض أن ﻿1x⋅ex\frac{1}{x}\cdot e^xx1​⋅ex﻿ ﻿q(x)=q\left(x\right)=q(x)=﻿ ، أوجد ﻿q′(x)q'\left(x\right)q′(x)﻿ .

﻿q′(x)=−1x2⋅ex+1x⋅exq'\left(x\right)=-\frac{1}{x^2}\cdot e^x+\frac{1}{x}\cdot e^xq′(x)=−x21​⋅ex+x1​⋅ex﻿

﻿q′(x)=−1x2⋅ex−1x⋅exq'\left(x\right)=-\frac{1}{x^2}\cdot e^x-\frac{1}{x}\cdot e^xq′(x)=−x21​⋅ex−x1​⋅ex﻿

﻿q′(x)=−1x2⋅ex+1x⋅eq'\left(x\right)=-\frac{1}{x^2}\cdot e^x+\frac{1}{x}\cdot e^{ }q′(x)=−x21​⋅ex+x1​⋅e﻿

أوجد مشتقة الدالة ﻿ex⋅sin⁡xe^x\cdot\sin xex⋅sinx﻿ ﻿h(x)=h\left(x\right)=h(x)=﻿ .

﻿h′(x)=ex⋅sin⁡x+ex⋅cos⁡xh'\left(x\right)=e^x\cdot\sin x+e^x\cdot\cos xh′(x)=ex⋅sinx+ex⋅cosx﻿

﻿h′(x)=ex⋅sin⁡x−ex⋅cos⁡xh'\left(x\right)=e^x\cdot\sin x-e^x\cdot\cos xh′(x)=ex⋅sinx−ex⋅cosx﻿

﻿h′(x)=ex⋅sin⁡x+cos⁡xh'\left(x\right)=e^x\cdot\sin x+\cos xh′(x)=ex⋅sinx+cosx﻿

إجابة غير صحيحة

رائع!

إذا كان $\sqrt{x}\cdot\tan x$ $u\left(x\right)=$ ، أوجد $u'\left(x\right)$ .

$u'\left(x\right)=\frac{1}{2\sqrt{x}}\cdot\tan x+\sqrt{x}\cdot\sec^2x$

$u'\left(x\right)=12\sqrt{x}\cdot\tan x+\sqrt{x}\cdot\sec^2x$

$u'\left(x\right)=\frac{1}{2\sqrt{x}}\cdot\tan x-\sqrt{x}\cdot\sec^2$

افترض أن $x\cdot\cos x\cdot\ln x$ $w\left(x\right)=$ . أوجد $w'\left(x\right)$ .

$w'\left(x\right)=\cos x\cdot\ln x-x\cdot\sin x\cdot\ln x+x\cdot\cos x\cdot\frac{1}{x}$

$w'\left(x\right)=\cos x\cdot\ln x-x\cdot\sin x\cdot\ln x-x\cdot\cos x\cdot\frac{1}{x}$

$w'\left(x\right)=\cos x\cdot\ln x+x\cdot\sin x\cdot\ln x+x\cdot\cos x\cdot\frac{1}{x}$

إذا كان $\sin x-e^x$ $v\left(x\right)=$ ، فاحسب $v'\left(x\right)$ .

$v'\left(x\right)=\cos x\cdot e^x+\sin x\cdot e^{ }$

$v'\left(x\right)=\cos x\cdot e^x-\sin x\cdot e^x$

$v'\left(x\right)=\cos x\cdot e^x+\sin x\cdot e^x$

افترض أن $x^3\cdot\ln x$ $m\left(x\right)=$ ، احسب m'(x).

$m'\left(x\right)=3x^2.\ln x-x^2.\frac{1}{x}$

$m'\left(x\right)=3x^2.\ln x+x^2.\frac{1}{x}$

$m'\left(x\right)=3x^2.\ln x+\frac{1}{x}$

$p'\left(x\right)=\frac{1}{2\sqrt{x}}\cdot\cos x+\sin x\cdot\sqrt{x}$

$p'\left(x\right)=\frac{1}{2\sqrt{x}}\cdot\cos x-\sin x$

$p'\left(x\right)=\frac{1}{2\sqrt{x}}\cdot\cos x-\sin x\cdot\sqrt{x}$

أوجد مشتقة $\ln x\cdot e^{-x}$ $t\left(x\right)=$ .

$t'\left(x\right)=\frac{1}{x}\cdot e^{-x}-\ln x\cdot e^{-x}$

$t'\left(x\right)=\frac{1}{x}\cdot e^{-x}+\ln x\cdot e^{-x}$

$t'\left(x\right)=\frac{1}{x}\cdot e^{-x}-\ln x\cdot e^x$

إذا كان $\cos x\cdot\ln x$ $r\left(x\right)=$ ، فأوجد $r'\left(x\right)$ .

$r'\left(x\right)=\sin x\cdot\ln x-\frac{\cos x}{x}$

$r'\left(x\right)=\sin x\cdot\ln x+\frac{\cos x}{x}$

$r'\left(x\right)=\sin x\cdot\ln x+\cos x$

احسب مشتقة $x\cdot e^x\cdot\sin x$ $s\left(x\right)=$ .

$s'\left(x\right)=e^x\cdot\sin x-x\cdot e^x\cdot\cos x-x\cdot e^x\sin x$

$s'\left(x\right)=e^x\cdot\sin x+x\cdot e^x\cdot\cos-x\cdot e^x\sin x$

$s'\left(x\right)=e^x\cdot\sin x+x\cdot e^x\cdot\cos x+x\cdot e^x\sin x$

افترض أن $\frac{1}{x}\cdot e^x$ $q\left(x\right)=$ ، أوجد $q'\left(x\right)$ .

$q'\left(x\right)=-\frac{1}{x^2}\cdot e^x+\frac{1}{x}\cdot e^x$

$q'\left(x\right)=-\frac{1}{x^2}\cdot e^x-\frac{1}{x}\cdot e^x$

$q'\left(x\right)=-\frac{1}{x^2}\cdot e^x+\frac{1}{x}\cdot e^{ }$

أوجد مشتقة الدالة $e^x\cdot\sin x$ $h\left(x\right)=$ .

$h'\left(x\right)=e^x\cdot\sin x+e^x\cdot\cos x$

$h'\left(x\right)=e^x\cdot\sin x-e^x\cdot\cos x$

$h'\left(x\right)=e^x\cdot\sin x+\cos x$