تمارين (2-3)

1- جد باستخدام القواعد مشتقة كل من الدوال الآتية:

- $f\left(x\right)=x^3-4x^2+x-1 , x=1$

$\begin{array}{rl}& \overline{f}\left(x\right)=3x^2-8x+1\\ & \overline{f}\left(1\right)=3(1{)}^2-8(1)+1=3-8+1=-4\end{array}$

- $f\left(x\right)=(4-x)(x^2+3) , x=2$

$\begin{array}{rl}& \overline{f}\left(x\right)=(4-x)\left(2x\right)+(x^2+3)(-1)\\ & \overline{f}\left(2\right)=(4-2)\left(4\right)+(4+3)(-1)=8-7=1\end{array}$

- $f\left(x\right)=\frac{4-5x}{x^2+x+1} , x=-1$

$\begin{array}{rl}& \overline{f}\left(x\right)=\frac{(x^2+x+1)(-5)-(4-5x)(2x+1)}{{(x^2+x+1)}^2}\\ & \overline{f}\left(x\right)=\frac{(1-1+1)(-5)-(4-5(-1\left)\right)(-2+1)}{(1-1+1{)}^2}\\ & \overline{f}\left(x\right)=\frac{(-5)-(4+5)(-1)}{1}=-5+9=4\end{array}$

- $f\left(x\right)=\frac{1}{\sqrt{2x+1}} , x=0$

$\begin{array}{rl}& f\left(x\right)=(2x+1{)}^{-\frac{1}{2}}\\ & \overline{f}\left(x\right)=-\frac{1}{2}(2x+1{)}^{-\frac{3}{2}}(2)=-(2x+1{)}^{-\frac{3}{2}}\\ & \overline{f}\left(x\right)=\frac{-1}{(2x+1{)}^{\frac{3}{2}}}\Rightarrow \overline{f}\left(0\right)=\frac{-1}{(0+1{)}^{\frac{3}{2}}}=-1\end{array}$

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

$\begin{array}{rl}& \overline{f}\left(x\right)=1+\frac{(x^2+2)\left(0\right)-3\left(2x\right)}{{(x^2+2)}^2}\\ & \overline{f}(-1)=1+\frac{-3(-2)}{(1+2{)}^2}=1+\frac{6}{9}\Rightarrow \overline{f}(-1)=1+\frac{2}{3}=\frac{5}{3}\end{array}$

2- إذا كانت $\overline{f}\left(x\right)={(x^2-3)}^4$ جد $\overline{\overline{f}}\left(x\right),\overline{f}\left(x\right)$

المشتقة الأولى

$\begin{array}{rl}& \overline{f}\left(x\right)=4{(x^2-3)}^3\left(2x\right)=8x{(x^2-3)}^3\\ & \overline{f}\left(2\right)=8\left(2\right)(4-3{)}^3=16(1{)}^3=16\\ & \overline{\overline{f}}\left(x\right)=8x\left(3\right){(x^2-3)}^2\left(2x\right)+{(x^2-3)}^3\left(8\right)\\ & \overline{\overline{f}}\left(2\right)=8\left(2\right)\left(3\right)(4-3{)}^2(4)+(4-3{)}^3\left(8\right)\\ & \overline{\overline{f}}\left(2\right)=48\left(1{)}^2\right(4)+(1{)}^3\left(8\right)=192+8=200\end{array}$

3- إذا كانت $f\left(x\right)={(x^3+3x^2-3)}^{\frac{3}{2}}$ جد $\overline{f}\left(x\right)$

$\begin{array}{rl}\overline{f}\left(x\right)& =\frac{3}{2}{(x^3+3x^2-3)}^{\frac{1}{2}}(3x^2+6x)\\ \overline{f}\left(2\right)& =\frac{3}{2}(8+12-3{)}^{\frac{1}{2}}(12+12)\\ \overline{f}\left(2\right)& =\frac{3}{2}(20-3{)}^{\frac{1}{2}}(12)=36(17{)}^{\frac{1}{2}}\\ & =36\sqrt{17}\end{array}$