المشتقة الضمنية

إذا كانت $y=f\left(x\right)$ دالة بدلالة $x$ فعند اشتقاق معادلة بدلالة $x , y$ بالنسبة إلى $x$ نضيف $y\text{'}$ أو $\frac{dy}{dx}$ بعد ل مشتقة ل $y$ وتستخدم المشتقة الضمنية عندما يكون قيمة أس $y$ أكبر من واحد كما يأتي:

(1)- إذا كانت $y=\cos 2x$ فجد $\frac{d^{4}y}{d{x}^4}$

$\begin{array}{rl}& \frac{dy}{dx}=-2\sin 2x , \frac{d^{2}y}{d{x}^2}=-2\cos 2x\cdot \left(2\right)=-4\cos 2x\\ & \frac{d^{3}y}{d{x}^3}=8\sin 2x\Rightarrow \frac{d^{4}y}{d{x}^4}=16\cos 2x\end{array}$

(2)- إذا كانت $y^2+x^2=1$ فأثبت أن $y\frac{d^{2}y}{d{x}^2}+{\left(\frac{dy}{dx}\right)}^2+1=0$

نشتق العلاقة المعطاة اشتقاقاً ضمنياً بالنسبة إلى $x$

$\begin{array}{rl}& 2y\frac{dy}{dx}+2x=0]÷2\\ & y\frac{dy}{dx}+x=0\Rightarrow y\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}\cdot \frac{dy}{dx}+1=0\Rightarrow y\frac{d^{2}y}{d{x}^2}+{\left(\frac{dy}{dx}\right)}^2+1=0\end{array}$

(3)- لتكن $xy-13=0$ حيث $x\ne 0 , y\ne 0$ فجد المشتقة الثانية:

$\begin{array}{rl}& x\frac{dy}{dx}+y-0=0\Rightarrow x\frac{dy}{dx}+y=0\\ & x\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}\cdot \left(1\right)+\frac{dy}{dx}=0\Rightarrow x\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}+\frac{dy}{dx}=0\Rightarrow x\frac{d^{2}y}{d{x}^2}+2\frac{dy}{dx}=0\Rightarrow x{y}^{\prime \prime }=-2y^{\mathrm{\prime }}\\ & \therefore y^{\prime \prime }=\frac{-2y^{\mathrm{\prime }}}{x}\end{array}$

(4)- إذا كانت $y={\cos }^{2}x-{\sin }^{2}x$ أثبت أن $y^{\prime \prime }=-4{\cos }^{2}x$

$$\begin{gathered} y={\cos }^{2}x-{\sin }^{2}x\Rightarrow y=\cos 2x \\ {y}^{\mathrm{\prime }}=-\sin 2x.\left(2\right)=-2\sin 2x\Rightarrow y^{\prime \prime }=-2\cos 2x\cdot \left(2\right)=-4\cos 2x \end{gathered}$$

(5)- جد $y^{\prime \prime \prime \prime }$ للدالة $f\left(x\right)=x^5+2x^3+3x+1$

$\begin{array}{rl}& y\text{'}=5x^4+6x^2+3\\ & y\text{'}\text{'}=20x^3+12x\\ & y^{\prime \prime \mathrm{\prime }}=60x^2+12\\ & y^{\prime \prime \prime \prime }=120x\end{array}$

(6)- إذا كانت $y={\sin }^{4}x$ أثبت أن $\frac{d^{2}y}{d{x}^2}+16y=12{\sin }^{2}x$

$\begin{array}{rl}& y={\sin }^{4}x\Rightarrow y=[\sin x{]}^4\\ & \frac{dy}{dx}=4[\sin x{]}^3\cdot [\cos x]\\ & \frac{d^{2}y}{d{x}^2}=4[\sin x{]}^3\cdot (-\sin x)+(\cos x)\cdot (12{\sin }^{2}x)\cdot (\cos x)\\ & \frac{d^{2}y}{d{x}^2}=-4{\sin }^{4}x+12{\sin }^{2}x{\cos }^{2}x\\ & \frac{d^{2}y}{d{x}^2}=-4{\sin }^{4}x+12{\sin }^{2}x(1-{\sin }^{2}x)\\ & \frac{d^{2}y}{d{x}^2}=-4{\sin }^{4}x+12{\sin }^{2}x-12{\sin }^{4}x\\ & \frac{d^{2}y}{d{x}^2}=-16{\sin }^{4}x+12{\sin }^{2}x\\ & \therefore \frac{d^{2}y}{d{x}^2}+16{\sin }^{4}x=12{\sin }^{2}x\Rightarrow \frac{d^{2}y}{d{x}^2}+16y=12{\sin }^{2}x\end{array}$

(7)- جد $y^{\mathrm{\prime }}$ للدوال الآتية:

$f\left(x\right)=\sin (2x^2+x+3)$

$f\left(x\right)=\sin (2x^2+x+3)\Rightarrow y^{\mathrm{\prime }}=(4x+1)\cos (2x^2+x+3)$

$f\left(x\right)={\cot }^3\sqrt{x}$

$f\left(x\right)={\cot }^3\sqrt{x}\Rightarrow y^{\mathrm{\prime }}=\frac{-1}{\underset{\mathrm{الزاوية} \mathrm{مشتقة}}{\underbrace{3\sqrt[3]{x^2}}}}{\csc }^2\sqrt[3]{x}$

$f\left(x\right)=\sec \frac{1}{x}$

$f\left(x\right)=\sec \frac{1}{x}\Rightarrow y^{\mathrm{\prime }}=\frac{-1}{x^2}\sec \frac{1}{x}\tan \frac{1}{x}$

$f\left(x\right)=\cos x\tan x$

$f\left(x\right)=\cos x\tan x\Rightarrow y^{\mathrm{\prime }}=\cos x\cdot {\sec }^{2}x+\tan x\cdot (-\sin x)=\cos x\cdot {\sec }^{2}x-\tan x\cdot \sin x$

(8)- جد $y^{\mathrm{\prime }}$ للدوال الآتية:

إذا كانت الدالة مرفوعة لقوة نضع القوة خارج القوس الكبير ثم نشتق الدالة حسب الدالة القوسية.

$f\left(x\right)={\sin }^3(\pi x^2+3x+2)$

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

$f\left(x\right)={\cos }^{-4}{x}^2$

$f\left(x\right)={\cos }^{-4}{x}^2\Rightarrow f\left(x\right)={[\cos x^2]}^{-4}\Rightarrow y^{\mathrm{\prime }}=-4{[\cos x^2]}^{-5}\cdot (\underset{\mathrm{الزاوية} \mathrm{مشتقة}}{\underbrace{-2x}}\sin x^2)$

$f\left(x\right)={\sec }^{\frac{2}{3}}{x}^2$

$$\begin{gathered} f\left(x\right)={\sec }^{\frac{2}{3}}{x}^2\Rightarrow f\left(x\right)={[\sec x^2]}^{\frac{2}{3}} \\ {f}^{\mathrm{\prime }}\left(x\right)=\frac{2}{3}{[\sec x^2]}^{\frac{-1}{3}}\cdot (\sec x^2\tan x^2)\left(\underset{\mathrm{الزاوية} \mathrm{مشتقة}}{\underbrace{2x}}\right)=\frac{4x}{3}{[\sec x^2]}^{\frac{-1}{3}}(\sec x^2\tan x^2) \end{gathered}$$

(9)- جد $y^{\mathrm{\prime }}$ للدوال الآتية:

$\sin \left(xy\right)=x^2+3y^{\mathrm{\prime }}$

$$\begin{gathered} \sin \left(xy\right)=x^2+3y\Rightarrow \cos xy[x{y}^{\mathrm{\prime }}+y(1\left)\right]=2x+3y^{\mathrm{\prime }} \\ \begin{array}{rl}& x{y}^{\mathrm{\prime }}\cos xy+y\cos xy=2x+3y^{\mathrm{\prime }}\\ & x{y}^{\mathrm{\prime }}\cos xy-3y^{\mathrm{\prime }}=2x-y\cos xy\\ & y^{\mathrm{\prime }}(x\cos xy-3)=2x-y\cos xy\Rightarrow y^{\mathrm{\prime }}=\frac{2x-y\cos xy}{x\cos xy-3}\end{array} \end{gathered}$$

$\sqrt{\tan x}=2y^2+x$

$\frac{{\sec }^{2}x}{2\sqrt{\tan x}}=4y{y}^{\mathrm{\prime }}+1\Rightarrow \frac{{\sec }^{2}x}{2\sqrt{\tan x}}-1=4y{y}^{\mathrm{\prime }}\Rightarrow y^{\mathrm{\prime }}=\frac{\frac{{\sec }^{2}x}{2\sqrt{\tan x}}-1}{4y}$

(10)- جد $y^{\mathrm{\prime }}$ لما يأتي: $y=(\sin x+\cos x{)}^2$

$$\begin{gathered} \frac{dy}{dx}=2(\sin x+\cos x{)}^1(\underset{\mathrm{القوس} \mathrm{داخل} \mathrm{مشتقة}}{\underbrace{\cos x-\sin x}})\Rightarrow \frac{dy}{dx}=2(\sin x+\cos x\left)\right(\cos x-\sin x) \\ \begin{array}{rl}& \frac{dy}{dx}=2({\sin }^{2}x-{\cos }^{2}x)\left(\cos 2x={\cos }^{2}x-{\sin }^{2}x\right)\\ & \frac{dy}{dx}=2\cos 2x\end{array} \end{gathered}$$