Question

Alguém poderia me ajudar nessas contas de derivada?

Answer 1

a)

$\frac{\partial f(x)}{\partial x}=e^x+4x^{4-1}\\\\\frac{\partial f(x)}{\partial x}=e^x+4x^3$

b)

$\frac{\partial y(x)}{\partial x}=\left(3(x^2-2x)^2.(2x^1-2)\right).(3+5x)+5.(x^2-2x)^3\\\\\frac{\partial y(x)}{\partial x}=\left((x^2-2x)^2.(6x-6)\right).(3+5x)+5.(x^2-2x)^3$

Eu deixaria nessa forma, no entanto, caso queira fazer as multiplicacoes e simplificacoes, fica:

$x^2(35x^4-162x^3+210x^2-16x-72)$

c)

$\frac{\partial f(x)}{\partial x}=\frac{3.(x^2-3x+4)-(2x-3).(3x-7)}{(x^2-3x+4)^2}\\\\\\frac{\partial f(x)}{\partial x}=frac{(3x^2-9x+12)-(6x^2-23x+21)}{(x^2-3x+4)^2}\\\\\\frac{\partial f(x)}{\partial x}=frac{-3x^2+14x-9}{(x^2-3x+4)^2}$

d)

$\frac{\partial f(x)}{\partial x}=\frac{\partial }{\partial x}\left(2.x^{-3}-7.x^{-5}\right)\\\\\frac{\partial f(x)}{\partial x}=-3.2x^{-3-1}-5.(-7)x^{-5-1}\\\\\frac{\partial f(x)}{\partial x}=-6x^{-4}+35x^{-6}\\\\\frac{\partial f(x)}{\partial x}=-\frac{6}{x^4}+\frac{35}{x^6}$