Atrisiniet f^prime(x)=-ax^2-2bx+ac/(x^2+c)^2

$$f ^ { \prime } ( x ) = \frac { - a x ^ { 2 } - 2 b x + a c } { ( x ^ { 2 } + c ) ^ { 2 } }$$

Atrast a

$$\left\{\begin{matrix}a=-\frac{2bx}{x^{2}-c}\text{, }&|c|\neq x^{2}\\a\in \mathrm{R}\text{, }&b=0\text{ and }c=x^{2}\text{ and }x\neq 0\end{matrix}\right,$$

Lineāra vienādojuma risināšanas darbības

Reiziniet vienādojuma abas puses ar \left(x^{2}+c\right)^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(f)x\left(x^{2}+c\right)^{2}=\left(-a\right)x^{2}-2bx+ac$$

Lietojiet Ņūtona binomu \left(a+b\right)^{2}=a^{2}+2ab+b^{2}, lai izvērstu \left(x^{2}+c\right)^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(f)x\left(\left(x^{2}\right)^{2}+2x^{2}c+c^{2}\right)=\left(-a\right)x^{2}-2bx+ac$$

Lai pakāpi kāpinātu citā pakāpē, sareiziniet kāpinātājus. Sareiziniet 2 un 2, lai iegūtu 4.

$$\frac{\mathrm{d}}{\mathrm{d}x}(f)x\left(x^{4}+2x^{2}c+c^{2}\right)=\left(-a\right)x^{2}-2bx+ac$$

Izmantojiet distributīvo īpašību, lai reizinātu \frac{\mathrm{d}}{\mathrm{d}x}(f)x ar x^{4}+2x^{2}c+c^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(f)x^{5}+2\frac{\mathrm{d}}{\mathrm{d}x}(f)cx^{3}+\frac{\mathrm{d}}{\mathrm{d}x}(f)xc^{2}=\left(-a\right)x^{2}-2bx+ac$$

Mainiet puses tā, lai visi mainīgie locekļi atrastos pa kreisi.

$$\left(-a\right)x^{2}-2bx+ac=\frac{\mathrm{d}}{\mathrm{d}x}(f)x^{5}+2\frac{\mathrm{d}}{\mathrm{d}x}(f)cx^{3}+\frac{\mathrm{d}}{\mathrm{d}x}(f)xc^{2}$$

Pievienot 2bx abās pusēs.

$$\left(-a\right)x^{2}+ac=\frac{\mathrm{d}}{\mathrm{d}x}(f)x^{5}+2\frac{\mathrm{d}}{\mathrm{d}x}(f)cx^{3}+\frac{\mathrm{d}}{\mathrm{d}x}(f)xc^{2}+2bx$$

Pārkārtojiet locekļus.

$$-ax^{2}+ac=x^{5}\frac{\mathrm{d}}{\mathrm{d}x}(f)+2cx^{3}\frac{\mathrm{d}}{\mathrm{d}x}(f)+xc^{2}\frac{\mathrm{d}}{\mathrm{d}x}(f)+2bx$$

Savelciet visus locekļus, kuros ir a.

$$\left(-x^{2}+c\right)a=x^{5}\frac{\mathrm{d}}{\mathrm{d}x}(f)+2cx^{3}\frac{\mathrm{d}}{\mathrm{d}x}(f)+xc^{2}\frac{\mathrm{d}}{\mathrm{d}x}(f)+2bx$$

Vienādojums ir standarta formā.

$$\left(c-x^{2}\right)a=2bx$$

Daliet abas puses ar -x^{2}+c.

$$\frac{\left(c-x^{2}\right)a}{c-x^{2}}=\frac{2bx}{c-x^{2}}$$

Dalīšana ar -x^{2}+c atsauc reizināšanu ar -x^{2}+c.

$$a=\frac{2bx}{c-x^{2}}$$

Atrast b

$$\left\{\begin{matrix}b=\frac{a\left(c-x^{2}\right)}{2x}\text{, }&x\neq 0\text{ and }c\neq -x^{2}\\b\in \mathrm{R}\text{, }&a=0\text{ and }x=0\text{ and }c\neq 0\end{matrix}\right,$$

Lineāra vienādojuma risināšanas darbības

Reiziniet vienādojuma abas puses ar \left(x^{2}+c\right)^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(f)x\left(x^{2}+c\right)^{2}=\left(-a\right)x^{2}-2bx+ac$$

Lietojiet Ņūtona binomu \left(a+b\right)^{2}=a^{2}+2ab+b^{2}, lai izvērstu \left(x^{2}+c\right)^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(f)x\left(\left(x^{2}\right)^{2}+2x^{2}c+c^{2}\right)=\left(-a\right)x^{2}-2bx+ac$$

Lai pakāpi kāpinātu citā pakāpē, sareiziniet kāpinātājus. Sareiziniet 2 un 2, lai iegūtu 4.

$$\frac{\mathrm{d}}{\mathrm{d}x}(f)x\left(x^{4}+2x^{2}c+c^{2}\right)=\left(-a\right)x^{2}-2bx+ac$$

Izmantojiet distributīvo īpašību, lai reizinātu \frac{\mathrm{d}}{\mathrm{d}x}(f)x ar x^{4}+2x^{2}c+c^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(f)x^{5}+2\frac{\mathrm{d}}{\mathrm{d}x}(f)cx^{3}+\frac{\mathrm{d}}{\mathrm{d}x}(f)xc^{2}=\left(-a\right)x^{2}-2bx+ac$$

Mainiet puses tā, lai visi mainīgie locekļi atrastos pa kreisi.

$$\left(-a\right)x^{2}-2bx+ac=\frac{\mathrm{d}}{\mathrm{d}x}(f)x^{5}+2\frac{\mathrm{d}}{\mathrm{d}x}(f)cx^{3}+\frac{\mathrm{d}}{\mathrm{d}x}(f)xc^{2}$$

Atņemiet \left(-a\right)x^{2} no abām pusēm.

$$-2bx+ac=\frac{\mathrm{d}}{\mathrm{d}x}(f)x^{5}+2\frac{\mathrm{d}}{\mathrm{d}x}(f)cx^{3}+\frac{\mathrm{d}}{\mathrm{d}x}(f)xc^{2}-\left(-a\right)x^{2}$$

Atņemiet ac no abām pusēm.

$$-2bx=\frac{\mathrm{d}}{\mathrm{d}x}(f)x^{5}+2\frac{\mathrm{d}}{\mathrm{d}x}(f)cx^{3}+\frac{\mathrm{d}}{\mathrm{d}x}(f)xc^{2}-\left(-a\right)x^{2}-ac$$

Reiziniet -1 un -1, lai iegūtu 1.

$$-2bx=\frac{\mathrm{d}}{\mathrm{d}x}(f)x^{5}+2\frac{\mathrm{d}}{\mathrm{d}x}(f)cx^{3}+\frac{\mathrm{d}}{\mathrm{d}x}(f)xc^{2}+ax^{2}-ac$$

Vienādojums ir standarta formā.

$$\left(-2x\right)b=ax^{2}-ac$$

Daliet abas puses ar -2x.

$$\frac{\left(-2x\right)b}{-2x}=\frac{a\left(x^{2}-c\right)}{-2x}$$

Dalīšana ar -2x atsauc reizināšanu ar -2x.

$$b=\frac{a\left(x^{2}-c\right)}{-2x}$$

Daliet a\left(x^{2}-c\right) ar -2x.

$$b=-\frac{ax}{2}+\frac{ac}{2x}$$