Правило дифференцирования сложной функции:
(u(v))' = u'(v)*v'
1)
![f(x)=(x^3-x^5+6)^7=[v=x^3-x^5+6; u=v^7]=(v^7)'*v'=7v^{7-1}*(x^3-x^5+6)'=7v^6*(3x^{3-1}-5x^{5-1})=7(x^3-x^5+6)^6*(3x^{2}-5x^{4})](https://tex.z-dn.net/?f=f%28x%29%3D%28x%5E3-x%5E5%2B6%29%5E7%3D%5Bv%3Dx%5E3-x%5E5%2B6%3B+u%3Dv%5E7%5D%3D%28v%5E7%29%27%2Av%27%3D7v%5E%7B7-1%7D%2A%28x%5E3-x%5E5%2B6%29%27%3D7v%5E6%2A%283x%5E%7B3-1%7D-5x%5E%7B5-1%7D%29%3D7%28x%5E3-x%5E5%2B6%29%5E6%2A%283x%5E%7B2%7D-5x%5E%7B4%7D%29)
2)
![f(x)=((5-3x^3+x^4)^6)=[v=5-3x^3+x^4; u=v^6]=(v^6)'*v'=6v^{6-1}*(5-3x^3+x^4)'=6v^5*(-3*3x^{3-1}+4x^{4-1})=6(5-3x^3+x^4)^5*(-9x^2+4x^3)](https://tex.z-dn.net/?f=f%28x%29%3D%28%285-3x%5E3%2Bx%5E4%29%5E6%29%3D%5Bv%3D5-3x%5E3%2Bx%5E4%3B+u%3Dv%5E6%5D%3D%28v%5E6%29%27%2Av%27%3D6v%5E%7B6-1%7D%2A%285-3x%5E3%2Bx%5E4%29%27%3D6v%5E5%2A%28-3%2A3x%5E%7B3-1%7D%2B4x%5E%7B4-1%7D%29%3D6%285-3x%5E3%2Bx%5E4%29%5E5%2A%28-9x%5E2%2B4x%5E3%29)
3)
![f(x)=(3\sqrt{x}+2x)^5=5(3\sqrt{x}+2x)^{5-1}*(3\sqrt{x}+2x)'=\\\\=5(3\sqrt{x}+2x)^{4}*(3x^{\frac{1}{2}}+2x)'=5(3\sqrt{x}+2x)^{4}*(3*\frac{1}{2}x^{\frac{1}{2}-1}+2x^{1-1})=5(3\sqrt{x}+2x)^{4}*(\frac{3}{2}x^{-\frac{1}{2}}+2x^{1-1})=5(3\sqrt{x}+2x)^{4}*(\frac{3}{2\sqrt{x}}+2x^{0})=5(3\sqrt{x}+2x)^{4}*(\frac{3}{2\sqrt{x}}+2)](https://tex.z-dn.net/?f=f%28x%29%3D%283%5Csqrt%7Bx%7D%2B2x%29%5E5%3D5%283%5Csqrt%7Bx%7D%2B2x%29%5E%7B5-1%7D%2A%283%5Csqrt%7Bx%7D%2B2x%29%27%3D%5C%5C%5C%5C%3D5%283%5Csqrt%7Bx%7D%2B2x%29%5E%7B4%7D%2A%283x%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%2B2x%29%27%3D5%283%5Csqrt%7Bx%7D%2B2x%29%5E%7B4%7D%2A%283%2A%5Cfrac%7B1%7D%7B2%7Dx%5E%7B%5Cfrac%7B1%7D%7B2%7D-1%7D%2B2x%5E%7B1-1%7D%29%3D5%283%5Csqrt%7Bx%7D%2B2x%29%5E%7B4%7D%2A%28%5Cfrac%7B3%7D%7B2%7Dx%5E%7B-%5Cfrac%7B1%7D%7B2%7D%7D%2B2x%5E%7B1-1%7D%29%3D5%283%5Csqrt%7Bx%7D%2B2x%29%5E%7B4%7D%2A%28%5Cfrac%7B3%7D%7B2%5Csqrt%7Bx%7D%7D%2B2x%5E%7B0%7D%29%3D5%283%5Csqrt%7Bx%7D%2B2x%29%5E%7B4%7D%2A%28%5Cfrac%7B3%7D%7B2%5Csqrt%7Bx%7D%7D%2B2%29)
Решение приложено у снимку:
![f(x) = \frac{2x}{ {x}^{2} - 1 }](https://tex.z-dn.net/?f=f%28x%29+%3D++%5Cfrac%7B2x%7D%7B+%7Bx%7D%5E%7B2%7D+-+1+%7D+)
В общем случае,
функция определена при всех х,
кроме тех, при которых х²-1=0
то есть при х≠ ±1
x€(-∞, -1)V(-1; 1)V(1;+∞)
Если у Вас, как написано х<0
то x€(-∞, -1)V(-1; 0)