Question

determine a derivada pela definições de limite, sendo dado: f(x) = 5^2-1
f(x) = 4x-3

Answer 1

Olá


Derivada por definição

$\displaystyle \mathsf{ \lim_{h \to 0} ~ \frac{f(x+h)-f(x)}{h} }\\\\\\\text{Dado f(x)}\\\\\\\mathsf{f(x)=5x^2-1}\\\\\\\mathsf{ \lim_{h \to 0} ~ \frac{(5\cdot (x+h)^2-1)-(5x^2-1)}{h} }\\\\\\\mathsf{ \lim_{h \to 0} ~ \frac{(5\cdot(x^2+2xh+h^2)-1)-(5x^2-1)}{h} }\\\\\\\mathsf{ \lim_{h \to 0} ~ \frac{\diagup\!\!\!\!\!5x^2+10xh+5h^2-\diagup\!\!\!1-\diagup\!\!\!\!\!5x^2+\diagup\!\!\!\!1}{h} }\\\\\\\mathsf{ \lim_{h \to 0} ~ \frac{10xh+5h^2}{h} }$

Colocando o 'h' em evidencia

$\displaystyle \mathsf{ \lim_{h \to 0} ~ \frac{\diagup\!\!\!\!h(10x+5h)}{\diagup\!\!\!\!h} }\\\\\\\mathsf{ \lim_{h \to 0} ~ (10x+5h) ~=~10x+5\cdot 0~=~\boxed{10x} }}~~\Leftarrow \text{Resposta}$


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$\displaystyle \mathsf{ \lim_{h \to 0} ~ \frac{f(x+h)-f(x)}{h} }\\\\\\\text{Dado f(x)}\\\\\\\mathsf{f(x)=4x-3}\\\\\\\mathsf{ \lim_{h \to 0} ~ \frac{(4(x+h)-3)-(4x-3)}{h} }\\\\\\\mathsf{ \lim_{h \to 0} ~ \frac{\diagup\!\!\!\!4x+4h-\diagup\!\!\!\!3-\diagup\!\!\!\!\!4x+\diagup\!\!\!\!\!3}{h} }\\\\\\\mathsf{ \lim_{h \to 0} ~ \frac{4\diagup\!\!\!\!h}{\diagup\!\!\!\!h} ~=~\boxed{4}}$