Question

решить ФНП, 2 семестр тех. ВУЗа. Для функции $F = [f(x^2-4y)]^{g(x^3-y)}$ найти все частные производные

Answer 1

$F'_x =[f(x^2-4y)]^{g(x^3-y)}\left[3x^2g'_x(x^3-y) \ln [f(x^2-4y)] + \frac{2x *g(x^3-y)f'_x(x^2-4y)}{f(x^2-4y)}\right]\\ \\F'_y =[f(x^2-4y)]^{g(x^3-y)}\left[-g'_y(x^3-y) \ln [f(x^2-4y)]-\frac{4*g(x^3-y)f'_y(x^2-4y)}{f(x^2-4y)}\right]$

Пошаговое объяснение:

прологарифмируем обе части

$\ln F=\ln [f(x^2-4y)]^{g(x^3-y}) \\ \\ \ln F=g(x^3-y) \ln [f(x^2-4y)] \\ \\ 1) \ \frac{F'_x}{F} =[g'_x(x^3-y)*(x^3-y)'_x* \ln [f(x^2-4y)]+ \\ \\ +g(x^3-y)*(\ln [f(x^2-4y)])'_x*f'_x(x^2-4y)*(x^2-4y)'_x] \\ \\ \frac{F'_x}{F} =[g'_x(x^3-y) *3x^2*\ln [f(x^2-4y)]+ \\ \\ +g(x^3-y)* \frac{1}{f(x^2-4y)}*f'_x(x^2-4y)*2x] \\ \\$

$F'_x =F*\left[3x^2g'_x(x^3-y) \ln [f(x^2-4y)] + \frac{2x *g(x^3-y)f'_x(x^2-4y)}{f(x^2-4y)}\right] \\ \\ F'_x =[f(x^2-4y)]^{g(x^3-y)}\left[3x^2g'_x(x^3-y) \ln [f(x^2-4y)] + \frac{2x *g(x^3-y)f'_x(x^2-4y)}{f(x^2-4y)}\right]$

$2) \ \frac{F'_y}{F} =[g'_y(x^3-y)*(x^3-y)'_y* \ln [f(x^2-4y)]+ \\ \\ +g(x^3-y)*(\ln [f(x^2-4y)])'_y*f'_y(x^2-4y)*(x^2-4y)'_y] \\ \\ \frac{F'_y}{F} =[g'_y(x^3-y)*(-1)* \ln [f(x^2-4y)]+ \\ \\ +g(x^3-y)*\frac{1}{f(x^2-4y)}*f'_y(x^2-4y)*(-4)] \\ \\ F'_y =F*\left[-g'_y(x^3-y) \ln [f(x^2-4y)]-\frac{4*g(x^3-y)f'_y(x^2-4y)}{f(x^2-4y)}\right] \\ \\ F'_y =[f(x^2-4y)]^{g(x^3-y)}\left[-g'_y(x^3-y) \ln [f(x^2-4y)]-\frac{4*g(x^3-y)f'_y(x^2-4y)}{f(x^2-4y)}\right]$