Дифференцирование функций 10.11.2025

1. Параметрически заданная функция

Вид

$$\{\begin{array}{l}y=y(t)\\ x=x(t)\end{array}$$

Вычисление производной

$$\{\begin{array}{l}{y}_x^{\prime }=\frac{y^{\prime }(t)}{x^{\prime }(t)}\\ x=x(t)\end{array}$$

Примеры РК2

$\{\begin{array}{l}y=t^3\\ x=2t-1\end{array}$

$$\{\begin{array}{l}{y}_t^{\prime }=3t^2\\ x_t=2\end{array};\{\begin{array}{l}{y}_x=1,5t^2\\ x=2t-1\end{array}$$

$\{\begin{array}{l}y=e^{2t}\\ x=e^t\end{array}$

$$\{\begin{array}{l}{y}_t^{\prime }=e^{2t}\\ x_t^{\prime }=e^t\end{array};\{\begin{array}{l}{y}_x^{\prime }=e^t\\ x=e^t\end{array}$$

$\{\begin{array}{l}y=\frac{\ln t}{t}\\ x=t\ln t\end{array}$ $t=1$

$$\{\begin{array}{l}{y}_t^{\prime }=\frac{1-\ln t}{t^2}\\ x_t=\ln t+1\end{array};\{\begin{array}{l}{y}_x^{\prime }=\frac{1-\ln t}{t^2(1+\ln t)}\\ x=t\ln t\end{array};y_x^{\prime }(t=1)=\frac{1}{1}=1$$

$\{\begin{array}{l}y=e^t\sin t\\ x=e^t\cos t\end{array}$ $t=\frac{\pi }{4}$

$$\{\begin{array}{l}{y}_t^{\prime }=e^t\sin t+e^t\cos t\\ x_t^{\prime }=e^t\cos t-e^t\sin t\end{array};\{\begin{array}{l}{y}_x^{\prime }=\frac{\sin t+\cos t}{\sin t-\cos t}\\ x=e^t\cos t\end{array};y_x^{\prime }(t=\frac{\pi }{4})=\frac{\sqrt{2}}{0}=\infty$$

2. Логарифмическая функция

Без него не обойтись при дифференцировании функций вида $y=f(x{)}^{g(x)}$

$$y=e^{\ln f(x{)}^{g(x)}}=e^{g(x)\ln f(x)}=>y^{\prime }=(e^{g(x)\ln f(x)}{)}^{\prime }=e^{g(x)\ln f(x)}\cdot (g(x)\cdot \ln f(x){)}^{\prime }$$

Формула в этом случае: $y^{\prime }=y(\ln y{)}^{\prime }$

Примеры

$y=x^{\sin x}$

$$y^{\prime }=x^{\sin x}(\cos x\ln x+\frac{\sin x}{x})$$

$y=(1-\frac{1}{x}{)}^x$

$$y^{\prime }=(1-\frac{1}{x}{)}^x(ln(1-\frac{1}{x})+\frac{x}{1+\frac{1}{x}})$$

$y=\frac{\sqrt{x-1}}{\sqrt[3]{(x+2{)}^2}\sqrt{(x+3{)}^3}}$

$$y^{\prime }=\frac{\sqrt{x-1}}{\sqrt[3]{(x+2{)}^2}\cdot \sqrt{(x+3{)}^3}}\cdot (\ln \frac{\sqrt{x-1}}{\sqrt[3]{(x+2{)}^2}\cdot \sqrt{(x+3{)}^3}}{)}^{\prime }=$$ $$=y\cdot (\ln \sqrt{x-1}-\ln (\sqrt[3]{(x+2{)}^2}\cdot \sqrt{(x+3{)}^3}))=y\cdot (\frac{0.5}{x-1}-\frac{2}{3x+6}-\frac{1,5}{x+3})$$

3. Неявно заданная функция

Вид: $F(x,y)=0$

$(F(x,y){)}_x^{\prime }=0_x^{\prime }=0=>\text{Найдём }{y}^{\prime }(x)$

Примеры

$x^2+y^2=1,y_x^{\prime }$

$$x^2+y^2-1=0=>(x^2+y^2-1{)}_x^{\prime }=0=>2x+2y\cdot y_x^{\prime }-0=0=>y_x^{\prime }=-\frac{x}{y}$$

$\sqrt{x}+\sqrt{y}=\sqrt{a}$

$$\sqrt{x}+\sqrt{y}-\sqrt{a}=0=>(\sqrt{x}+\sqrt{y}-\sqrt{a}{)}_x^{\prime }=0=>\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}\cdot y_x^{\prime }=0=>y_x^{\prime }=-\frac{\sqrt{x}}{\sqrt{y}}$$

$xy=\arctan \frac{x}{y}$

$$(xy-\arctan \frac{x}{y}{)}_x^{\prime }=0=>y+x{y}_x^{\prime }-\frac{1}{1+(\frac{x}{y}{)}^2}\cdot \frac{y-x{y}^{\prime }}{y^2}=0=>$$ $$=>y+x{y}_x^{\prime }-\frac{y-x{y}_x^{\prime }}{y^2+x^2}=0=>y+x{y}_x^{\prime }-\frac{y}{y^2+x^2}+\frac{x{y}_x^{\prime }}{y^2+x^2}=0=>$$ $$=>y_x^{\prime }(x+\frac{x}{y^2+x^2})+y-\frac{y}{y^2+x^2}=>y_x^{\prime }=\frac{\frac{y}{y^2+x^2}-y}{x+\frac{x}{y^2+x^2}}$$

$y{e}^y=e^{x+1}$

$$(y{e}^y-e^{x+1}{)}_x^{\prime }=0=>y_x^{\prime }{e}^y+y{e}^y{y}_x^{\prime }-e^{x+1}=0=>y_x^{\prime }=\frac{e^{x+1}}{e^y+y{e}^y}$$

4. Производные высших порядков

$y^{\prime \prime \prime \prime }=(y^{\prime \prime \prime }{)}^{\prime }=((y^{\prime \prime }{)}^{\prime }{)}^{\prime }=(((y^{\prime }{)}^{\prime }{)}^{\prime }{)}^{\prime }$

Примеры

$y=\ln \sqrt[3]{1+x^2}$

$$y^{\prime \prime }=(y^{\prime }{)}^{\prime }=(\frac{2x}{3+3x^2}{)}^{\prime }=\frac{2x}{(3+3x^2{)}^2}$$

$\{\begin{array}{l}x=\ln t\\ y=t^3\end{array}$

$$\{\begin{array}{l}{x}_t^{\prime }=\frac{1}{t}\\ y_t^{\prime }=3t^2\end{array};\{\begin{array}{l}x=\ln t\\ y_x^{\prime }=3t^3\end{array};\{\begin{array}{l}{x}_t^{\prime }=\frac{1}{t}\\ (y_t^{\prime }{)}_t^{\prime }=9t^2\end{array};\{\begin{array}{l}x=\ln t\\ y_{xx}^{\prime \prime }=9t^3\end{array}$$

$y=x+\arctan (y)$

$$(y-x-\arctan (y){)}_x^{\prime }=0;y_x^{\prime }=\frac{1}{y^2}+1y_{xx}^{\prime \prime }=-\frac{2}{y^3}$$

$x^4+y^4-xy=0$

$$(x^4+y^4-xy{)}_x^{\prime }=0=>y_x^{\prime }=\frac{y-4x^3}{4y^3-x}=>y_{xx}^{\prime \prime }=\frac{(y_x^{\prime }-12x)(4y^3-x)-(4y_x^{\prime }-1)(y-4x^3)}{(4y^3-x{)}^2}$$