Question

Por favor me ajudem

Derive as Funções:

Answer 1

Resposta:

a) 9x² + 3eˣx² + 6eˣx - 5eˣ - 5

b) eˣ/2x² + 1

c) 3x + 2ln(x) + 2/(3x + 2)²

Explicação passo-a-passo:

a)

$\frac{d}{dx}\left(\left(x+e^x\right)\left(3x^2-5\right)\right)\\=\frac{d}{dx}\left(x+e^x\right)\left(3x^2-5\right)+\frac{d}{dx}\left(3x^2-5\right)\left(x+e^x\right)\\=\left(1+e^x\right)\left(3x^2-5\right)+6x\left(x+e^x\right)\\=9x^2+3e^xx^2+6e^xx-5e^x-5$

b)

$\frac{d}{dg}\left(gx\frac{e^x}{2x^3+x}\right)\\=x\frac{e^x}{2x^3+x}\frac{d}{dg}\left(g\right)\\=x\frac{e^x}{2x^3+x}\cdot \:1\\=\frac{e^x}{2x^2+1}$

c)

$\frac{d}{dx}\left(\frac{x^2\ln \left(x\right)}{3x^2+2x}\right)\\=\frac{\frac{d}{dx}\left(x^2\ln \left(x\right)\right)\left(3x^2+2x\right)-\frac{d}{dx}\left(3x^2+2x\right)x^2\ln \left(x\right)}{\left(3x^2+2x\right)^2}\\=\frac{\left(2x\ln \left(x\right)+x\right)\left(3x^2+2x\right)-\left(6x+2\right)x^2\ln \left(x\right)}{\left(3x^2+2x\right)^2}\\=\frac{3x+2\ln \left(x\right)+2}{\left(3x+2\right)^2}$