Question

ajuda urgente, bastante duvida nessas.

Answer 1

$\boxed{\dfrac{\mathrm{d}}{\mathrm{d}x}\bigg(f(x)g(x)\bigg)=f'(x)g(x)+f(x)g'(x)}$

$\boxed{\dfrac{\mathrm{d}}{\mathrm{d}x}\bigg(f(x)g(x)h(x)\bigg)=f'(x)g(x)h(x)+f(x)g'(x)h(x)+f(x)g(x)h'(x)}$

$\boxed{\dfrac{\mathrm{d}}{\mathrm{d}x}\bigg(\dfrac{f(x)}{g(x)}\bigg)=\dfrac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}}$

a)

$f(x)=2\bigg(x+\dfrac{1}{x}\bigg)(x-3)$

$\dfrac{\mathrm{d}f}{\mathrm{d}x}=2\bigg[(x-3)\dfrac{\mathrm{d}}{\mathrm{d}x}\bigg(x+\dfrac{1}{x}\bigg)+\bigg(x+\dfrac{1}{x}\bigg)\dfrac{\mathrm{d}}{\mathrm{d}x}(x-3)\bigg]$

$\dfrac{\mathrm{d}f}{\mathrm{d}x}=2\bigg[(x-3)\bigg(1-\dfrac{1}{x^2}\bigg)+\bigg(x+\dfrac{1}{x}\bigg)(1)\bigg]$

$\dfrac{\mathrm{d}f}{\mathrm{d}x}=2\bigg[x-\dfrac{1}{x}-3+\dfrac{3}{x^2}+x+\dfrac{1}{x}\bigg]$

$\boxed{\dfrac{\mathrm{d}f}{\mathrm{d}x}=2\bigg[2x-3+\dfrac{3}{x^2}\bigg]}$

b)

$f(x)=\dfrac{x^2-2}{3x+1}$

$\dfrac{\mathrm{d}f}{\mathrm{d}x}=\dfrac{(3x+1)\frac{\mathrm{d}}{\mathrm{d}x}(x^2-2)-(x^2-2)\frac{\mathrm{d}}{\mathrm{d}x}(3x+1)}{(3x+1)^2}$

$\dfrac{\mathrm{d}f}{\mathrm{d}x}=\dfrac{(3x+1)(2x)-(x^2-2)(3)}{(3x+1)^2}$

$\dfrac{\mathrm{d}f}{\mathrm{d}x}=\dfrac{6x^2+2x-3x^2+6}{(3x+1)^2}$

$\boxed{\dfrac{\mathrm{d}f}{\mathrm{d}x}=\dfrac{3x^2+2x+6}{(3x+1)^2}}$

c)

$f(x)=(x+2)(2x^2-2x+3)(x^3-x+2)$

$\dfrac{\mathrm{d}f}{\mathrm{d}x}=(2x^2-2x+3)(x^3-x+2)\dfrac{\mathrm{d}}{\mathrm{d}x}(x+2)+\\\\ +(x+2)(x^3-x+2)\dfrac{\mathrm{d}}{\mathrm{d}x}(2x^2-2x+3)+\\\\ +(x+2)(2x^2-2x+3)\dfrac{\mathrm{d}}{\mathrm{d}x}(x^3-x+2)$

$\dfrac{\mathrm{d}f}{\mathrm{d}x}=(2x^5-2x^4+x^3+6x^2-7x+6)(1)+\\\\ +(x^4+2x^3-x^2+4)(4x-2)+(2x^3+2x^2-x+6)(3x^2-1)$

$\dfrac{\mathrm{d}f}{\mathrm{d}x}=2x^5-2x^4+x^3+6x^2-7x+6+\\\\ +4x^5+6x^4-8x^3+2x^2+16x-8+6x^5+6x^4-5x^3+16x^2+x-6$

$\boxed{\dfrac{\mathrm{d}f}{\mathrm{d}x}=12x^5+10x^4-12x^3+24x^2+10x-8}$