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أتدرب وأحل المسائل

قاعدة السلسلة الأسئلة (1 - 24)

أجد مشتقة كل اقتران ممّا يأتي:

(1) f(x) = e^{4x + 2}

f '(x) = 4e^{4x + 2}

(2) f(x) = 50e^{2x - 10}

f '(x) = 100e^{2x - 10}

(3) f(x) = cos (x² – 3x – 4)

f '(x) = - (2x – 3) sin (x² – 3x – 4)

f '(x) = (3 – 2x) sin (x² – 3x – 4)

(4) f(x) = 10x² $e^{-x^2}$

f '(x) = (10x²) (-2x$e^{-x^2}$) + ($e^{-x^2}$) (20x) = 20x$e^{-x^2}$ (1 – x²)

(5) f(x) = $\sqrt{\frac{x + 1}{x}}$

f (x) = $\sqrt{\frac{x + 1}{x}}$ = $\sqrt{1 + \frac{1}{x}}$  

f '(x) = $\frac{-{\displaystyle \frac{1}{x^2}}}{\sqrt[2]{1 + {\displaystyle \frac{1}{x}}}}$ = - $\frac{-1}{{2x}^2 \sqrt{1 + {\displaystyle \frac{1}{x}}} }$

(6) f(x) = x² tan $\frac{1}{x}$

f '(x) = (x²) (- $\frac{1}{x^2}$ sec² $\frac{1}{x}$) + (tan $\frac{1}{x}$) (2x) 

f '(x) = -sec² $\frac{1}{x}$ +  2x tan $\frac{1}{x}$

(7) f(x) = 3x – 5 cos (πx)²

f '(x) = 3 + 5(2) (πx) (π) sin (πx)² = 3 + 10π²x sin (πx)²

(8) f(x) = ln ($\frac{1 + e^x}{1 - e^x}$)

f (x) = ln ($\frac{1 + e^x}{1 - e^x}$) = ln (1 + e^x) – ln (1 – e^x)

f '(x) = $\frac{e^x}{1 + e^x}$ + $\frac{e^x}{1 - e^x}$ = $\frac{2e^x}{1 - {e^2}^x}$

(9) f(x) = (ln x)⁴

f '(x) = $\frac{4}{x}$ (ln x)³

(10) f(x) = sin $\sqrt[3]{x}$ + $\sqrt[3]{\sin x}$

f '(x) = $\frac{1}{3\sqrt[3]{x^2}}$ cos $\sqrt[3]{x}$ + $\frac{\cos x}{3\sqrt[3]{{\sin }^2 x}}$

(11) f(x) = $\sqrt[5]{x^2+ 8x}$

f '(x) = $\frac{2x + 8}{5\sqrt[5]{{(x^2 + 8x)}^4}}$

(12) f(x) = $\frac{3^{2x}}{x}$

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

(13) f(x) = 2^-x cos πx

f '(x) = (2^-x) (-π sin πx) + (cos πx) (-ln 2)2^-x

= - π 2^-x sin πx – 2^-x (cos πx) ln 2

(14) f(x) = $\frac{10 {\log }_4 x}{x}$

f '(x) = $\frac{ {\displaystyle \frac{10x}{x \ln 4}}- 10 {\log }_4 x}{x^2}$ = $\frac{ {\displaystyle \frac{10}{\ln 4}}- 10 {\log }_4 x}{x^2}$

(15) f(x) = ($\frac{\sin x}{1 + \cos x}$)²

f '(x) = 2 ($\frac{\sin x}{1 + \cos x}$)¹ x $\frac{(1 + \cos x) (\cos x) - (\sin x) (-\sin x)}{{(1 + \cos x)}^2}$ 

         = 2 x $\frac{\sin x}{1 + \cos x}$ x $\frac{1}{1 + \cos x}$

         = $\frac{2 \sin x}{{(1 + \cos x)}^2}$

(16) f(x) = log₃ (1 + x ln x)

f '(x) = $\frac{\left(x\right) \left({\displaystyle \frac{1}{x}}\right) + (\ln x) \left(1\right)}{(\ln 3) (1 + x \ln x)}$ = $\frac{1+ \ln x}{(\ln 3) (1 + x \ln x)}$

(17) f(x) = e^{sin 2x} + sin (e^2x)

f '(x) = 2e^{sin 2x} cos 2x + 2e^2x cos (e^2x)

(18) f(x) = tan⁴ (sec (cos x))

f '(x) = 4(tan (sec(cos x)))³ sec² (sec(cos x)) x sec(cos x) tan(cos x) x (-sin x)  

  = -4 tan³ (sec(cos x)) sec² (sec(cos x)) sec(cos x) tan(cos x) sin x  

 

أجد معادلة المماس لكل اقتران ممّا يأتي عند قيمة x المعطاة:

(19) $f\left(x\right)=4e^{-0.5x^2} , x=-2$

$\begin{array}{rl}& f\left(x\right)=4e^{-0.5x^2}\\ & f(-2)=4e^{-0.5(-2{)}^2}=\frac{4}{e^2}\\ & f^{\mathrm{\prime }}\left(x\right)=-4x{e}^{-0.5x^2}\end{array}$

ميل المماس هو:

$m=f^{\mathrm{\prime }}(-2)=-4(-2)e^{-0.5(-2{)}^2}=\frac{8}{e^2}$

معادلة المماس هي:

$y-\frac{4}{e^2}=\frac{8}{e^2}(x+2)\to y=\frac{8}{e^2}x+\frac{20}{e^2}$

(20) $f\left(x\right)=x+\cos 2x , x=0$

$\begin{array}{rl}& f\left(x\right)=x+cos2x\\ & f\left(0\right)=0+cos\left(0\right)=1\\ & f^{\mathrm{\prime }}\left(x\right)=1-2sin 2x\end{array}$

ميل المماس هو:

$m=f^{\mathrm{\prime }}\left(0\right)=1-2sin2\left(0\right)=1$

معادلة المماس هي:

$y-1=1(x-0)\to y=x+1$

(21) $f\left(x\right)=2^x , x=0$

$\begin{array}{rl}& f\left(x\right)=2^x\\ & f\left(0\right)=2^0=1\\ & f^{\mathrm{\prime }}\left(x\right)=(ln2)2^x\end{array}$

ميل المماس هو:

$m=f^{\mathrm{\prime }}\left(0\right)=(ln2)2^0=ln2$

معادلة المماس هي:

$y-1=(ln2)(x-0)\to y=(ln2)x+1$

(22) $f\left(x\right)=\sqrt{x+1}\sin \frac{\pi x}{2} , x=3$

$\begin{array}{rl}& f\left(x\right)=\sqrt{x+1}sin\frac{\pi x}{2}\\ & f\left(3\right)=2sin\frac{3\pi }{2}=-2\\ & f^{\mathrm{\prime }}\left(x\right)=\left(\sqrt{x+1}\right)(\frac{\pi }{2}cos\frac{\pi x}{2})+(sin\frac{\pi x}{2})\left(\frac{1}{2\sqrt{x+1}}\right)\end{array}$

ميل المماس هو:

$m=f^{\mathrm{\prime }}\left(3\right)=\left(2\right)\left(0\right)+(-1)\left(\frac{1}{4}\right)=-\frac{1}{4}$

معادلة المماس هي:

$y+2=-\frac{1}{4}(x-3)\to y=-\frac{1}{4}x-\frac{5}{4}$

 

(23) إذا كان: $A\left(x\right)=f\left(g\right(x\left)\right)$ ، وكان: $f(-2)=8, f^{\mathrm{\prime }}(-2)=4,f^{\mathrm{\prime }}\left(5\right)=3, g\left(5\right)=-2, g^{\mathrm{\prime }}\left(5\right)=6$ فأجد $A^{\mathrm{\prime }}\left(5\right)$ .

$\begin{array}{rl}{A}^{\mathrm{\prime }}\left(x\right)& =f^{\mathrm{\prime }}\left(g\right(x\left)\right)\times g^{\mathrm{\prime }}\left(x\right)\\ A^{\mathrm{\prime }}\left(5\right)& =f^{\mathrm{\prime }}\left(g\right(5\left)\right)\times g^{\mathrm{\prime }}\left(5\right)\\ & =f^{\mathrm{\prime }}(-2)\times 6\\ & =4\times 6=24\end{array}$

 

(24) إذا كان: $f\left(x\right)=\frac{x}{\sqrt{x^2+1}}$ ، فأثبت أنّ $f^{\mathrm{\prime }}\left(x\right)=\frac{1}{\sqrt{{(x^2+1)}^3}}$ .

$\begin{array}{rl}f\left(x\right)& =\frac{x}{\sqrt{x^2+1}}\\ f^{\mathrm{\prime }}\left(x\right)& =\frac{\left(\sqrt{x^2+1}\right)\left(1\right)-\left(x\right)\left(\frac{x}{\sqrt{x^2+1}}\right)}{x^2+1}\\ & =\frac{\left(\frac{x^2+1-x^2}{\sqrt{x^2+1}}\right)}{x^2+1}\\ & =\frac{1}{(x^2+1)\sqrt{x^2+1}}\\ & =\frac{1}{\sqrt{{(x^2+1)}^3}}\end{array}$