Resposta:
4. [tex]a)\:y'=-\frac{1}{kx^2}[/tex]
5. [tex]b)\:y'=1[/tex]
Explicação passo a passo:
Questão 4:
[tex]y=\frac{1}{kx}\\\\y'=\frac{d}{dx}\left(\frac{1}{kx}\right)\\\\=\frac{1}{k}\frac{d}{dx}\left(\frac{1}{x}\right)\\\\=\frac{1}{k}\frac{d}{dx}\left(x^{-1}\right)\\\\=\frac{1}{k}\left(-1\cdot \:x^{-1-1}\right)\\\\=-\frac{1}{kx^2}[/tex]
Questão 5:
[tex]y=\frac{x^2-1}{x+1},\:x\ne -1\\\\y'=\frac{d}{dx}\left(\frac{x^2-1}{x+1}\right)\\\\=\frac{d}{dx}\left(\frac{\left(x+1\right)\left(x-1\right)}{x+1}\right)\\\\=\frac{d}{dx}\left(x-1\right)\\\\=\frac{d}{dx}\left(x\right)-\frac{d}{dx}\left(1\right)\\\\\frac{d}{dx}\left(x\right)=1\\\\\frac{d}{dx}\left(1\right)=0\\\\=1-0\\\\=1[/tex]
