Question

DERIVADAS
Usando a definição (de derivadas), calcule a derivada das seguintes funções nos pontos dados:

Answer 1

segue na imagem a solução.

Answer 2

a)

$f'(2)=\lim\limits_{h\rightarrow0}\dfrac{f(2+h)-f(2)}{h}$

Vamos achar f(2 + h):

$f(2+h)=2(2+h)^{2}-3(2+h)+4\\f(2+h)=2(2^{2}+2\cdot2\cdot h+h^{2})-6-3h+4\\f(2+h)=2(4+4h+h^{2})-2-3h\\f(2+h)=2h^{2}+8h+8-3h-2\\f(2+h)=2h^{2}+5h+6$

Então:

$f'(2)=\lim\limits_{h\rightarrow0}\dfrac{f(2+h)-f(2)}{h}\\\\\\f'(2)=\lim\limits_{h\rightarrow0}\dfrac{2h^{2}+5h+6-6}{h}\\\\\\f'(2)=\lim\limits_{h\rightarrow0}\dfrac{2h^{2}+5h}{h}\\\\\\f'(2)=\lim\limits_{h\rightarrow0}\dfrac{h(2h+5)}{h}\\\\\\f'(2)=\lim\limits_{h\rightarrow0}(2h+5)\\\\\\f'(2)=2\cdot0+5\\\\\\\boxed{\boxed{f'(2)=5}}$

b)

$f'(1)=\lim\limits_{h\rightarrow0}\dfrac{f(1+h)-f(1)}{h}$

Achando f(1 + h):

$f(1+h)=\dfrac{3}{(1+h)^{2}}=\dfrac{3}{(1^{2}+2\cdot1\cdot h+h^{2})}=\dfrac{3}{h^{2}+2h+1}$

Logo:

$f'(1)=\lim\limits_{h\rightarrow0}\dfrac{f(1+h)-f(1)}{h}\\\\\\f'(1)=\lim\limits_{h\rightarrow0}\dfrac{(\frac{3}{h^{2}+2h+1})-3}{h}\\\\\\f'(1)=\lim\limits_{h\rightarrow0}\dfrac{(\frac{3-3(h^{2}+2h+1)}{h^{2}+2h+1})}{h}\\\\\\f'(1)=\lim\limits_{h\rightarrow0}\dfrac{3-3h^{2}-6h-3}{h(h^{2}+2h+1)}\\\\\\f'(1)=\lim\limits_{h\rightarrow0}\dfrac{-3h^{2}-6h}{h(h^{2}+2h+1)}$

Colocando h em evidência no numerador:

$f'(1)=\lim\limits_{h\rightarrow0}\dfrac{h(-3h-6)}{h(h^{2}+2h+1)}\\\\\\f'(1)=\lim\limits_{h\rightarrow0}\dfrac{-3h-6}{h^{2}+2h+1}\\\\\\f'(1)=\dfrac{-3\cdot0-6}{0^{2}+2\cdot0+1}\\\\\\\boxed{\boxed{f'(1)=-6}}$