Question

Calcule a derivada de cada uma das funções abaixo, pelo cálculo direto do limite da razão incremental $\lim_{h \to \ 0} \frac{f(x+h)-f(x)}{h}$.

(a) f(x) = 7x − 5

(b) f(x) = 4x² − 3x

Answer 1

Explicação passo-a-passo:

a)Temos que  $f(x)=7x-5$.

$\dfrac{df}{dx}=\displaystyle \lim_{h \to \ 0}\dfrac{f(x+h)-f(x)}{h} \\\\ \dfrac{df}{dx}=\displaystyle \lim_{h \to \ 0}\dfrac{7.(x+h)-5-(7x-5)}{h}\\\\ \dfrac{df}{dx}=\displaystyle \lim_{h \to \ 0}\dfrac{7x+7h-5-7x+5}{h}\\\\ \dfrac{df}{dx}=\displaystyle \lim_{h \to \ 0}\dfrac{7h}{h}\\\\ \dfrac{df}{dx}=\displaystyle \lim_{h \to \ 0}7\\\\ \dfrac{df}{dx}=7$

b)

$\dfrac{df}{dx}=\displaystyle \lim_{h \to \ 0}\dfrac{f(x+h)-f(x)}{h}\\\\\dfrac{df}{dx}=\displaystyle \lim_{h \to \ 0}\dfrac{4.(x+h)^{2}-3.(x+h) -(4x^{2} -3x)}{h}\\\\\dfrac{df}{dx}=\displaystyle \lim_{h \to \ 0}\dfrac{4.(x^{2} +2xh+h^{2})-3x-3h -4x^{2} +3x}{h}\\\\\dfrac{df}{dx}=\displaystyle \lim_{h \to \ 0}\dfrac{4.x^{2} +8xh+4h^{2}-3h -4x^{2} }{h}\\\\\dfrac{df}{dx}=\displaystyle \lim_{h \to \ 0}\dfrac{8xh+4h^{2}-3h }{h}\\\\\dfrac{df}{dx}=\displaystyle \lim_{h \to \ 0}\dfrac{h.(8x+4h-3)}{h}$

$\dfrac{df}{dx}=\displaystyle \lim_{h \to \ 0}8x+4h-3\\\\\dfrac{df}{dx}=8x+4.0-3\\\\\dfrac{df}{dx}=8x-3$

Answer 2

$\large\boxed{\begin{array}{l}\tt a)\\\displaystyle\sf f'(x)=\lim_{h \to 0}\dfrac{f(x+h)-f(x)}{h}\\\displaystyle\sf f'(x)= \lim_{h \to 0}\dfrac{7\cdot(x+h)-5-(7x-5)}{h}\\\displaystyle\sf f'(x)=\lim_{h \to 0}\dfrac{\backslash\!\!\!\!7x+7h-\backslash\!\!\!5-\backslash\!\!\!\!\!7x+\backslash\!\!\!5}{h}\\\displaystyle\sf f'(x)=\lim_{h \to 0}\dfrac{7\backslash\!\!\!h}{\backslash\!\!\!h}\longrightarrow f'(x)=\lim_{h \to 0}7=7\end{array}}$

$\large\boxed{\begin{array}{l}\tt b)\\\displaystyle\sf f'(x)= \lim_{h \to 0}\dfrac{f(x+h)-f(x)}{h}\\\displaystyle\sf f'(x)=\lim_{h \to 0}\dfrac{4\cdot(x+h)^2-3\cdot(x+h)-(4x^2-3x)}{h}\\\displaystyle\sf f'(x)=\lim_{h \to 0}\dfrac{\backslash\!\!\!4x^2+8hx+4h^2-\backslash\!\!\!3x-3h-\backslash\!\!\!4x^2+\backslash\!\!\!3x}{h}\\\displaystyle\sf f'(x)=\lim_{ h \to 0}\dfrac{\diagup\!\!\!h(8x+4h-3)}{\diagup\!\!\!h}\\\displaystyle\sf f'(x)=\lim_{h \to 0}8x+4h-3=8x-3\end{array}}$