Question

Derivada de primeira das funções :
a) f(x) = x³/3 + x²/2- x/4
b) f(x) = 9x²/2 + 12/x-3/x²
c) f(x) = 1/x² - 4√x³

Answer 1

Resposta:

• a) f(x) = x³/3 + x²/2 - x/4

$\sf f'(x)=\big(\frac{x^3}{3}+\frac{x^2}{2}-\frac{x}{4}\big)'$ ⇒ Derive cada termo separadamente.

$\sf f'(x)=\big(\frac{x^3}{3}\big)'+\big(\frac{x^2}{2}\big)'-\big(\frac{x}{4}\big)'$ ⇒ Aplique a regra: (cxⁿ)' = c.(xⁿ)'

$\sf f'(x)=\frac{1}{3}\cdot(x^3)'+\frac{1}{2}\cdot(x^2)'-\frac{1}{4}\cdot(x)'$ ⇒ Aplique a regra: (xⁿ)' = n.xⁿ⁻¹

$\sf f'(x)=\frac{1}{3}\cdot3\cdot x^{3-1}+\frac{1}{2}\cdot2\cdot x^{2-1}-\frac{1}{4}\cdot1\cdot x^{1-1}$

$\sf f'(x)=x^2+x^1-\frac{1}{4} x^{0}$

$\sf f'(x)=x^2+x^1-\frac{1}{4}\cdot1$

$\red{\sf f'(x)=x^2+x-\frac{1}{4}}$

 

Faça os mesmos procedimentos nas outras funções.

 

• b) f(x) = 9x²/2 + 12/x - 3/x²

$\sf f'(x)=\big(\frac{9x^2}{2}+\frac{12}{x}-\frac{3\,}{x^2}\big)'$

$\sf f'(x)=\big(\frac{9x^2}{2}\big)'+\big(\frac{12}{x}\big)'-\big(\frac{3\,}{x^2}\big)'$

$\sf f'(x)=\frac{9}{2}\cdot(x^2)'+12\cdot\big(\frac{1}{x}\big)'-3\cdot\big(\frac{1\,}{x^2}\big)'$

$\sf f'(x)=\frac{9}{2}\cdot(x^2)'+12\cdot(x^{-1})'-3\cdot(x^{-2})'$

$\sf f'(x)=\frac{9}{2}\cdot2\cdot x^{2-1}+12\cdot(-1\cdot x^{-1-1})-3\cdot(-2\cdot x^{-2-1})$

$\sf f'(x)=9x^1-12x^{-2}+6x^{-3}$

$\sf f'(x)=9x-12\cdot\frac{1}{x^2}+6\cdot\frac{1}{x^3}$

$\red{\sf f'(x)=9x-\frac{12}{x^2}+\frac{6\,}{x^3}}$

• c) f(x) = 1/x² - 4√x³

$\sf f'(x)=\big(\frac{1}{x^2}-4\sqrt{x^3}\big)'$

$\sf f'(x)=\big(\frac{1}{x^2}\big)'-\big(4\sqrt{x^3}\big)'$

$\sf f'(x)=\big(\frac{1}{x^2}\big)'-4\cdot\big(\sqrt{x^3}\big)'$

$\sf f'(x)=\big(\frac{1}{x^2}\big)'-4\cdot\big[(x^3)^{\frac{1}{2}}\big]'$

$\sf f'(x)=\big(\frac{1}{x^2}\big)'-4\cdot\big(x^{\frac{3}{2}}\big)'$

$\sf f'(x)=(x^{-2})'-4\cdot\big(x^{\frac{3}{2}}\big)'$

$\sf f'(x)=-2\cdot x^{-2-1}-4\cdot\frac{3}{2}\cdot x^{\frac{3}{2}-1}$

$\sf f'(x)=-2x^{-3}-2\cdot3\cdot x^{\frac{1}{2}}$

$\sf f'(x)=-2\cdot\frac{1}{x^3}-6\sqrt{x}$

$\red{\sf f'(x)=-\frac{2\,}{x^3}-6\sqrt{x}}$