Question

Derivada de
$y=(x^4+8)lnx$

Answer 1

$\\ Y = (x^4+8)lnx \\ \\ Y' = (x^4+8)'lnx + (x^4+8)lnx' \\ \\ y' = (4x^3+0)*lnx + (x^4+8)* \frac{1}{x} \\ \\ y' = 4x^3Lnx+ \frac{x^4+8}{x} \\ \\ y' = \frac{x*(4x^3Lnx) + x^4+8}{x} \\ \\ y' = \frac{4x^5Lnx + x^4+8}{x}$

Answer 2

$\sf \displaystyle y=\left(x^4+8\right)inx\\\\\\\frac{d}{dx}\left(\left(x^4+8\right)inx\right)\\\\\\=in\frac{d}{dx}\left(\left(x^4+8\right)x\right)\\\\\\=in\left(\frac{d}{dx}\left(x^4+8\right)x+\frac{d}{dx}\left(x\right)\left(x^4+8\right)\right)\\\\\\=in\left(4x^3x+1\cdot \left(x^4+8\right)\right)\\\\\\\to \boxed{\sf =i\left(8n+5nx^4\right)}$