Задача Кузнецов Дифференцирование 18-1

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Условие задачи[править]

Найти производную указанного порядка.

y=\left(2x^2-7 \right) \ln{(x-1)},\ y^{V}=?

Решение[править]

y'=\left(\left(2x^2-7 \right) \ln{(x-1)}\right)'= 4x\cdot \ln{(x-1)} + \left(2x^2-7 \right)\cdot \frac{1}{x-1} =
= 4x\cdot \ln{(x-1)} + \frac{2x^2-7}{x-1}
y''=\left(y'\right)'=\left(4x\cdot \ln{(x-1)} + \frac{2x^2-7}{x-1}\right)'=
= 4\ln{(x-1)} + 4x \cdot \frac{1}{x-1} + \frac{4x\cdot (x-1) - \left(2x^2-7\right)\cdot 1}{(x-1)^2} =
= 4\ln{(x-1)} + \frac{4x^2 -4x}{(x-1)^2} + \frac{4x^2-4x - 2x^2+7}{(x-1)^2} =
= 4\ln{(x-1)} + \frac{6x^2 -8x + 7}{(x-1)^2}
y'''=\left(y''\right)'=\left(4\ln{(x-1)} + \frac{6x^2 -8x + 7}{(x-1)^2}\right)'=
 = 4\cdot \frac{1}{x-1} + \frac{(12x-8)\cdot (x-1)^2 - \left(6x^2 -8x + 7\right)\cdot 2(x-1)}{(x-1)^4} =
 = \frac{4(x-1)^2}{(x-1)^3} + \frac{(12x-8)\cdot (x-1) - \left(6x^2 -8x + 7\right)\cdot 2}{(x-1)^3} =
 = \frac{4x^2-8x+4 + 12x^2 -12x-8x+8 - 12x^2 +16x - 14}{(x-1)^3} =
 = \frac{4x^2-12x-2}{(x-1)^3}
y^{(4)}=\left(y'''\right)'=\left(\frac{4x^2-12x-2}{(x-1)^3}\right)'=
 = \frac{(8x-12)\cdot (x-1)^3 - \left(4x^2-12x-2\right)\cdot 3(x-1)^2}{(x-1)^6}=
 = \frac{(8x-12)\cdot (x-1) - \left(4x^2-12x-2\right)\cdot 3}{(x-1)^4}=
 = \frac{8x^2-12x - 8x+12 - 12x^2+36x+6}{(x-1)^4} = \frac{-4x^2+16x+18}{(x-1)^4}
y^{(5)}=\left(y^{(4)}\right)'=\left(\frac{-4x^2+16x+18}{(x-1)^4}\right)'=
=\frac{(-8x+16)\cdot (x-1)^4 - \left(-4x^2+16x+18\right)\cdot 4(x-1)^3}{(x-1)^8}=
=\frac{(-8x+16)\cdot (x-1) - \left(-4x^2+16x+18\right)\cdot 4}{(x-1)^5}=
=\frac{-8x^2+16x +8x-16 +16x^2-64x-72}{(x-1)^5}=
=\frac{8x^2-40x-88}{(x-1)^5}=\frac{8\left(x^2-5x-11\right)}{(x-1)^5}