إذا كانت $y=x\sin x$ فبرهن أن: $y^{\left(4\right)}-y+4\cos x=0$

$\begin{array}{rl}& \frac{dy}{dx}=x\cdot \cos x+\sin x\left(1\right)\Rightarrow \frac{dy}{dx}=x\cos x+\sin x\\ & \frac{d^{2}y}{d{x}^2}=x\cdot (-\sin x)+\cos x\cdot \left(1\right)+\cos x=-x\sin x+2\cos x\\ & \frac{d^{3}y}{d{x}^3}=-x\cos x+\sin x(-1)-2\sin x=-x\cos x-\sin x-2\sin x\\ & \frac{d^{4}y}{d{x}^4}=-x(-\sin x)+\cos x\cdot (-1)-\cos x-2\cos x\\ & \frac{d^{4}y}{d{x}^4}=x\sin x-\cos x-\cos x-2\cos x=x\sin x-4\cos x\\ & \mathbf{\text{L.S.H}}=y^{\left(4\right)}-y+4\cos x=x\sin x-4\cos x-x\sin x+4\cos x=0=\mathbf{R}\cdot \mathbf{\text{S.H}}\end{array}$

تمارين (3-1)

تمارين (3-1)

(1)- جد $\frac{d^{2}y}{d{x}^2}$ مما يأتي:

$y=\sqrt{2-x} , \mathrm{\forall }x<2$

$\begin{array}{rl}& y=(2-x{)}^{\frac{1}{2}}\Rightarrow \frac{dy}{dx}=\frac{1}{2}(2-x{)}^{-\frac{1}{2}}\cdot (-1)=-\frac{1}{2}(2-x{)}^{-\frac{1}{2}}\\ & \frac{d^{2}y}{d{x}^2}=\frac{1}{4}(2-x{)}^{-\frac{3}{2}}\cdot (-1)=\frac{-1}{4(2-x{)}^{\frac{3}{2}}}\end{array}$

$y=\frac{2-x}{2+x} , x\ne -2$

$\begin{array}{rl}& \frac{dy}{dx}=\frac{(2+x)\cdot (-1)-(2-x)\cdot \left(1\right)}{(2+x{)}^2}=\frac{-2-x-2+x}{(2+x{)}^2}=\frac{-4}{(2+x{)}^2}=-4(2+x{)}^{-2}\\ & \frac{d^{2}y}{d{x}^2}=8(2+x{)}^{-3}\cdot (1)=\frac{8}{(2+x{)}^3}\end{array}$

$2xy-4y+5=0 , y\ne 0 , x\ne 2$

$\begin{array}{rl}& y(2x-4)=-5\Rightarrow y=\frac{-5}{(2x-4)}=-5(2x-4{)}^{-1}\\ & \frac{dy}{dx}=5(2x-4{)}^{-2}\cdot 2=10(2x-4{)}^{-2}\\ & \frac{d^{2}y}{d{x}^2}=-20(2x-4{)}^{-3}\cdot 2=\frac{-40}{(2x-4{)}^3}\end{array}$

(2)- جد $f\text{'}\text{'}\text{'}\left(1\right)$ لكل مما يأتي:

$f\left(x\right)=4\sqrt{6-2x} ,\hspace{0.17em}\mathrm{\forall }x<3$

$\begin{array}{rl}& f\left(x\right)=4(6-2x{)}^{\frac{1}{2}}\Rightarrow f\text{'}(x)=4\left(\frac{1}{2}\right)(6-2x{)}^{-\frac{1}{2}}\cdot (-2)=-4(6-2x{)}^{-\frac{1}{2}}\\ & f\text{'}\text{'}\left(x\right)=-4(-\frac{1}{2})(6-2x{)}^{-\frac{3}{2}}\cdot (-2)=-4(6-2x{)}^{-\frac{3}{2}}\\ & f\text{'}\text{'}\text{'}\left(x\right)=\frac{12}{2}(6-2x{)}^{-\frac{3}{2}-1}\cdot (-2)=-12(6-2x{)}^{-\frac{5}{2}}=\frac{-12}{(6-2x{)}^{\frac{5}{2}}}\\ & f\text{'}\text{'}\text{'}\left(1\right)=\frac{-12}{5}=\frac{-12}{5}=\frac{-12}{5}=\frac{-12}{22}=\frac{-3}{0}\end{array}$

$f\left(x\right)=\sin \pi x$

$$\begin{gathered} f\text{'}\left(x\right)=\cos \pi x.\left(\pi \right)=\pi \cos \pi x , f\text{'}\text{'}\left(x\right)=-\pi \sin \pi x.\left(\pi \right)=-{\pi }^2\sin \pi x \\ \begin{array}{rl}& f\text{'}\text{'}\text{'}\left(x\right)=-{\pi }^2\cos \pi x\left(\pi \right)=-{\pi }^3\cos \pi x\\ & f\text{'}\text{'}\text{'}\left(x\right)=-{\pi }^3\cos \pi \left(1\right)=-{\pi }^3(-1)={\pi }^3\end{array} \end{gathered}$$

$f\left(x\right)=\frac{3}{2-x} , x\ne 2$

$\begin{array}{rl}& f\left(x\right)=3(2-x{)}^{-1}\Rightarrow f\text{'}(x)=-3(2-x{)}^{-2}\cdot (-1)=3(2-x{)}^{-2}\\ & f\text{'}\text{'}\left(x\right)=-6(2-x{)}^{-3}\cdot (-1)=6(2-x{)}^{-3}\\ & f\text{'}\text{'}\text{'}\left(x\right)=-18(2-x{)}^{-4}(-1)=18(2-x{)}^{-4}=\frac{18}{(2-x{)}^4}\\ & \therefore f\text{'}\text{'}\text{'}\left(1\right)=\frac{18}{(2-1{)}^4}=18\end{array}$

(3)- إذا كانت $y=\tan x$ فبرهن أن $\frac{d^{2}y}{d{x}^2}=2y(1+y^2)$ حيث $x\ne \frac{(2n+1)\pi }{2} , n\in z$

$\begin{array}{rl}& \frac{dy}{dx}=[\sec x{]}^2\Rightarrow \frac{d^{2}y}{d{x}^2}=2[\sec x]\cdot \sec x\tan x=2\tan x{\sec }^{2}x\\ & \frac{d^{2}y}{d{x}^2}=2\tan x(1+{\tan }^{2}x)=2y(1+y^2)\end{array}$

(4)- إذا كانت $y=x\sin x$ فبرهن أن: $y^{\left(4\right)}-y+4\cos x=0$

$\begin{array}{rl}& \frac{dy}{dx}=x\cdot \cos x+\sin x\left(1\right)\Rightarrow \frac{dy}{dx}=x\cos x+\sin x\\ & \frac{d^{2}y}{d{x}^2}=x\cdot (-\sin x)+\cos x\cdot \left(1\right)+\cos x=-x\sin x+2\cos x\\ & \frac{d^{3}y}{d{x}^3}=-x\cos x+\sin x(-1)-2\sin x=-x\cos x-\sin x-2\sin x\\ & \frac{d^{4}y}{d{x}^4}=-x(-\sin x)+\cos x\cdot (-1)-\cos x-2\cos x\\ & \frac{d^{4}y}{d{x}^4}=x\sin x-\cos x-\cos x-2\cos x=x\sin x-4\cos x\\ & \mathbf{\text{L.S.H}}=y^{\left(4\right)}-y+4\cos x=x\sin x-4\cos x-x\sin x+4\cos x=0=\mathbf{R}\cdot \mathbf{\text{S.H}}\end{array}$