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الأسئلة

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الوقت

النتيجة

........................

إذا كان و. أوجد .

$\left(d+e\right)'\left(x\right)=d'\left(x\right)+e'\left(x\right)=\frac{2}{x^3}+12x^2$

$\left(d+e\right)'\left(x\right)=d'\left(x\right)+e'\left(x\right)=\frac{2}{x^3}-12x^2$

$\left(d+e\right)'\left(x\right)=d'\left(x\right)+e'\left(x\right)=\frac{2}{x^3}+12x^{ }$

أوجد مشتقة $e^x-\cos x+\ln x$ $c\left(x\right)=$ .

$c'\left(x\right)=e^x+\sin x+\frac{1}{x}$

$c'\left(x\right)=e^x+\sin x$

$c'\left(x\right)=e^x+\frac{1}{x}$

افترض أن الدالتين و . أوجد

$\left(f+g\right)'\left(x\right)=f'\left(x\right)+g'\left(x\right)=-\sin x+\frac{1}{x}$

$\left(f+g\right)'\left(x\right)=f'\left(x\right)+g'\left(x\right)=-\sin x-\frac{1}{x}$

$\left(f+g\right)'\left(x\right)=f'\left(x\right)-g'\left(x\right)=-\sin x-\frac{1}{x}$

إذا كان m(x)= 3cosx وn(x)= 2x²، احسب (m+n)'(x).

$\left(m+n\right)'\left(x\right)=m'\left(x\right)+n'\left(x\right)=-3\sin x+4x$

$\left(m+n\right)'\left(x\right)=m'\left(x\right)+n'\left(x\right)=3\sin x$

$\left(m+n\right)'\left(x\right)=m'\left(x\right)+n'\left(x\right)=3\sin x+x$

انظر الدالتين $\sqrt{x}$ $a\left(x\right)=$ و $3\ln x$ $b\left(x\right)=$ . أوجد $\left(a+b\right)'\left(x\right)$ .

$\left(a+b\right)'\left(x\right)=a'\left(x\right)+b'\left(x\right)=\frac{1}{2\sqrt{x}}+\frac{3}{x}$

$\left(a+b\right)'\left(x\right)=a'\left(x\right)+b'\left(x\right)=\frac{1}{2\sqrt{x}}$

$\left(a+b\right)'\left(x\right)=a'\left(x\right)+b'\left(x\right)=\frac{3}{x}$

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

$r'\left(x\right)=\frac{1}{x}$

$r'\left(x\right)=\frac{1}{x-\cos x}$

$r'\left(x\right)=\frac{1}{\cos x}$

إذا كان و ، احسب .

$\left(d+e\right)'\left(x\right)=d'\left(x\right)+e'\left(x\right)=-\frac{3}{x^4}+10^4$

$\left(d+e\right)'\left(x\right)=d'\left(x\right)+e'\left(x\right)=-\frac{3}{x^4}-10^4$

$\left(d+e\right)'\left(x\right)=d'\left(x\right)+e'\left(x\right)=\frac{3}{x^4}+10^4$

أوجد مشتقة $e^x+\tan x$ $q\left(x\right)=$ .

$q'\left(x\right)=e^x+\sec^2$

$q'\left(x\right)=e^x+\sec^2x$

$q'\left(x\right)=e^x+\sec^{ }$

أوجد مشتقة الدالة $h\left(x\right)=$ $\sqrt{x}-\ln x$ .

$h'\left(x\right)=\frac{1}{2\sqrt{x}}-\frac{1}{x}$

$h'\left(x\right)=\frac{1}{2\sqrt{x}}+\frac{1}{x}$

$h'\left(x\right)=\frac{1}{\sqrt{x}}+\frac{1}{x}$

بمعلومية أوجد

$\left(f+g\right)'\left(x\right)=f'\left(x\right)+g'\left(x\right)=4+e^x$

$\left(f+g\right)'\left(x\right)'=f'\left(x\right)+g'\left(x\right)=4$

$\left(f+g\right)'\left(x\right)'=f'\left(x\right)+g'\left(x\right)=e^x$

الأسئلة

الوقت

النتيجة

........................

إذا كان و. أوجد .

﻿(d+e)′(x)=d′(x)+e′(x)=2x3+12x2\left(d+e\right)'\left(x\right)=d'\left(x\right)+e'\left(x\right)=\frac{2}{x^3}+12x^2(d+e)′(x)=d′(x)+e′(x)=x32​+12x2﻿

﻿(d+e)′(x)=d′(x)+e′(x)=2x3−12x2\left(d+e\right)'\left(x\right)=d'\left(x\right)+e'\left(x\right)=\frac{2}{x^3}-12x^2(d+e)′(x)=d′(x)+e′(x)=x32​−12x2﻿

﻿(d+e)′(x)=d′(x)+e′(x)=2x3+12x\left(d+e\right)'\left(x\right)=d'\left(x\right)+e'\left(x\right)=\frac{2}{x^3}+12x^{ }(d+e)′(x)=d′(x)+e′(x)=x32​+12x﻿

أوجد مشتقة ﻿ex−cos⁡x+ln⁡xe^x-\cos x+\ln xex−cosx+lnx﻿ ﻿c(x)=c\left(x\right)=c(x)=﻿ .

﻿c′(x)=ex+sin⁡x+1xc'\left(x\right)=e^x+\sin x+\frac{1}{x}c′(x)=ex+sinx+x1​﻿

﻿c′(x)=ex+sin⁡xc'\left(x\right)=e^x+\sin xc′(x)=ex+sinx﻿

﻿c′(x)=ex+1xc'\left(x\right)=e^x+\frac{1}{x}c′(x)=ex+x1​﻿

افترض أن الدالتين و . أوجد

﻿(f+g)′(x)=f′(x)+g′(x)=−sin⁡x+1x\left(f+g\right)'\left(x\right)=f'\left(x\right)+g'\left(x\right)=-\sin x+\frac{1}{x}(f+g)′(x)=f′(x)+g′(x)=−sinx+x1​﻿

﻿(f+g)′(x)=f′(x)+g′(x)=−sin⁡x−1x\left(f+g\right)'\left(x\right)=f'\left(x\right)+g'\left(x\right)=-\sin x-\frac{1}{x}(f+g)′(x)=f′(x)+g′(x)=−sinx−x1​﻿

﻿(f+g)′(x)=f′(x)−g′(x)=−sin⁡x−1x\left(f+g\right)'\left(x\right)=f'\left(x\right)-g'\left(x\right)=-\sin x-\frac{1}{x}(f+g)′(x)=f′(x)−g′(x)=−sinx−x1​﻿

إذا كان m(x)= 3cosx وn(x)= 2x2﻿، احسب (m+n)'(x).

﻿(m+n)′(x)=m′(x)+n′(x)=−3sin⁡x+4x\left(m+n\right)'\left(x\right)=m'\left(x\right)+n'\left(x\right)=-3\sin x+4x(m+n)′(x)=m′(x)+n′(x)=−3sinx+4x﻿

﻿(m+n)′(x)=m′(x)+n′(x)=3sin⁡x\left(m+n\right)'\left(x\right)=m'\left(x\right)+n'\left(x\right)=3\sin x(m+n)′(x)=m′(x)+n′(x)=3sinx﻿

﻿(m+n)′(x)=m′(x)+n′(x)=3sin⁡x+x\left(m+n\right)'\left(x\right)=m'\left(x\right)+n'\left(x\right)=3\sin x+x(m+n)′(x)=m′(x)+n′(x)=3sinx+x﻿

انظر الدالتين ﻿x\sqrt{x}x​﻿ ﻿a(x)=a\left(x\right)=a(x)=﻿ و ﻿3ln⁡x3\ln x3lnx﻿ ﻿b(x)=b\left(x\right)=b(x)=﻿ . أوجد ﻿(a+b)′(x)\left(a+b\right)'\left(x\right)(a+b)′(x)﻿ .

﻿(a+b)′(x)=a′(x)+b′(x)=12x+3x\left(a+b\right)'\left(x\right)=a'\left(x\right)+b'\left(x\right)=\frac{1}{2\sqrt{x}}+\frac{3}{x}(a+b)′(x)=a′(x)+b′(x)=2x​1​+x3​﻿

﻿(a+b)′(x)=a′(x)+b′(x)=12x\left(a+b\right)'\left(x\right)=a'\left(x\right)+b'\left(x\right)=\frac{1}{2\sqrt{x}}(a+b)′(x)=a′(x)+b′(x)=2x​1​﻿

﻿(a+b)′(x)=a′(x)+b′(x)=3x\left(a+b\right)'\left(x\right)=a'\left(x\right)+b'\left(x\right)=\frac{3}{x}(a+b)′(x)=a′(x)+b′(x)=x3​﻿

إذا كان ﻿ln⁡x+sin⁡x\ln x+\sin xlnx+sinx﻿ ﻿r(x)=r\left(x\right)=r(x)=﻿ ، فأوجد r'(x).

﻿r′(x)=1xr'\left(x\right)=\frac{1}{x}r′(x)=x1​﻿

﻿r′(x)=1x−cos⁡xr'\left(x\right)=\frac{1}{x-\cos x}r′(x)=x−cosx1​﻿

﻿r′(x)=1cos⁡xr'\left(x\right)=\frac{1}{\cos x}r′(x)=cosx1​﻿

إذا كان و ، احسب .

﻿(d+e)′(x)=d′(x)+e′(x)=−3x4+104\left(d+e\right)'\left(x\right)=d'\left(x\right)+e'\left(x\right)=-\frac{3}{x^4}+10^4(d+e)′(x)=d′(x)+e′(x)=−x43​+104﻿

﻿(d+e)′(x)=d′(x)+e′(x)=−3x4−104\left(d+e\right)'\left(x\right)=d'\left(x\right)+e'\left(x\right)=-\frac{3}{x^4}-10^4(d+e)′(x)=d′(x)+e′(x)=−x43​−104﻿

﻿(d+e)′(x)=d′(x)+e′(x)=3x4+104\left(d+e\right)'\left(x\right)=d'\left(x\right)+e'\left(x\right)=\frac{3}{x^4}+10^4(d+e)′(x)=d′(x)+e′(x)=x43​+104﻿

أوجد مشتقة ﻿ex+tan⁡xe^x+\tan xex+tanx﻿ ﻿q(x)=q\left(x\right)=q(x)=﻿ .

﻿q′(x)=ex+sec⁡2q'\left(x\right)=e^x+\sec^2q′(x)=ex+sec2﻿

﻿q′(x)=ex+sec⁡2xq'\left(x\right)=e^x+\sec^2xq′(x)=ex+sec2x﻿

﻿q′(x)=ex+sec⁡q'\left(x\right)=e^x+\sec^{ }q′(x)=ex+sec﻿

أوجد مشتقة الدالة ﻿h(x)=h\left(x\right)=h(x)=﻿ ﻿x−ln⁡x\sqrt{x}-\ln xx​−lnx﻿ .

﻿h′(x)=12x−1xh'\left(x\right)=\frac{1}{2\sqrt{x}}-\frac{1}{x}h′(x)=2x​1​−x1​﻿

﻿h′(x)=12x+1xh'\left(x\right)=\frac{1}{2\sqrt{x}}+\frac{1}{x}h′(x)=2x​1​+x1​﻿

﻿h′(x)=1x+1xh'\left(x\right)=\frac{1}{\sqrt{x}}+\frac{1}{x}h′(x)=x​1​+x1​﻿

بمعلومية أوجد

﻿(f+g)′(x)=f′(x)+g′(x)=4+ex\left(f+g\right)'\left(x\right)=f'\left(x\right)+g'\left(x\right)=4+e^x(f+g)′(x)=f′(x)+g′(x)=4+ex﻿

﻿(f+g)′(x)′=f′(x)+g′(x)=4\left(f+g\right)'\left(x\right)'=f'\left(x\right)+g'\left(x\right)=4(f+g)′(x)′=f′(x)+g′(x)=4﻿

﻿(f+g)′(x)′=f′(x)+g′(x)=ex\left(f+g\right)'\left(x\right)'=f'\left(x\right)+g'\left(x\right)=e^x(f+g)′(x)′=f′(x)+g′(x)=ex﻿

إجابة غير صحيحة

رائع!

$\left(d+e\right)'\left(x\right)=d'\left(x\right)+e'\left(x\right)=\frac{2}{x^3}+12x^2$

$\left(d+e\right)'\left(x\right)=d'\left(x\right)+e'\left(x\right)=\frac{2}{x^3}-12x^2$

$\left(d+e\right)'\left(x\right)=d'\left(x\right)+e'\left(x\right)=\frac{2}{x^3}+12x^{ }$

أوجد مشتقة $e^x-\cos x+\ln x$ $c\left(x\right)=$ .

$c'\left(x\right)=e^x+\sin x+\frac{1}{x}$

$c'\left(x\right)=e^x+\sin x$

$c'\left(x\right)=e^x+\frac{1}{x}$

$\left(f+g\right)'\left(x\right)=f'\left(x\right)+g'\left(x\right)=-\sin x+\frac{1}{x}$

$\left(f+g\right)'\left(x\right)=f'\left(x\right)+g'\left(x\right)=-\sin x-\frac{1}{x}$

$\left(f+g\right)'\left(x\right)=f'\left(x\right)-g'\left(x\right)=-\sin x-\frac{1}{x}$

إذا كان m(x)= 3cosx وn(x)= 2x²، احسب (m+n)'(x).

$\left(m+n\right)'\left(x\right)=m'\left(x\right)+n'\left(x\right)=-3\sin x+4x$

$\left(m+n\right)'\left(x\right)=m'\left(x\right)+n'\left(x\right)=3\sin x$

$\left(m+n\right)'\left(x\right)=m'\left(x\right)+n'\left(x\right)=3\sin x+x$

انظر الدالتين $\sqrt{x}$ $a\left(x\right)=$ و $3\ln x$ $b\left(x\right)=$ . أوجد $\left(a+b\right)'\left(x\right)$ .

$\left(a+b\right)'\left(x\right)=a'\left(x\right)+b'\left(x\right)=\frac{1}{2\sqrt{x}}+\frac{3}{x}$

$\left(a+b\right)'\left(x\right)=a'\left(x\right)+b'\left(x\right)=\frac{1}{2\sqrt{x}}$

$\left(a+b\right)'\left(x\right)=a'\left(x\right)+b'\left(x\right)=\frac{3}{x}$

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

$r'\left(x\right)=\frac{1}{x}$

$r'\left(x\right)=\frac{1}{x-\cos x}$

$r'\left(x\right)=\frac{1}{\cos x}$

$\left(d+e\right)'\left(x\right)=d'\left(x\right)+e'\left(x\right)=-\frac{3}{x^4}+10^4$

$\left(d+e\right)'\left(x\right)=d'\left(x\right)+e'\left(x\right)=-\frac{3}{x^4}-10^4$

$\left(d+e\right)'\left(x\right)=d'\left(x\right)+e'\left(x\right)=\frac{3}{x^4}+10^4$

أوجد مشتقة $e^x+\tan x$ $q\left(x\right)=$ .

$q'\left(x\right)=e^x+\sec^2$

$q'\left(x\right)=e^x+\sec^2x$

$q'\left(x\right)=e^x+\sec^{ }$

أوجد مشتقة الدالة $h\left(x\right)=$ $\sqrt{x}-\ln x$ .

$h'\left(x\right)=\frac{1}{2\sqrt{x}}-\frac{1}{x}$

$h'\left(x\right)=\frac{1}{2\sqrt{x}}+\frac{1}{x}$

$h'\left(x\right)=\frac{1}{\sqrt{x}}+\frac{1}{x}$

$\left(f+g\right)'\left(x\right)=f'\left(x\right)+g'\left(x\right)=4+e^x$

$\left(f+g\right)'\left(x\right)'=f'\left(x\right)+g'\left(x\right)=4$

$\left(f+g\right)'\left(x\right)'=f'\left(x\right)+g'\left(x\right)=e^x$