Question

derivada de 1/(5x-3) no ponto x = 1

Answer 1

1/(5x - 3) , x = 1
1/(5×1 - 3) = 1/(5-3) = 1/2

Espero ter ajudado;

Answer 2

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Derivada no ponto

$\huge\boxed{\boxed{\boxed{\boxed{\displaystyle\sf f'(a)=\lim_{x \to a}\dfrac{f(x)-f(a)}{x-a}}}}}$

$\sf f(x)=\dfrac{1}{5x-3}\\\sf f(1)=\dfrac{1}{5\cdot1-3}=\dfrac{1}{5-3}=\dfrac{1}{2}\\\displaystyle\sf f'(1)=\lim_{x \to 1}\dfrac{\frac{1}{5x-3}-\frac{1}{2}}{x-1}\\\displaystyle\sf f'(1)=\lim_{x \to 1}\dfrac{\frac{2-1[5x-3]}{2\cdot(5x-3)}}{x-1}$

$\displaystyle\sf f'(1)=\lim_{x \to 1}\dfrac{\frac{2-5x+3}{2\cdot(5x-3)}}{x-1}\\\displaystyle\sf f'(1)=\lim_{x \to 1}\dfrac{1}{(x-1)}\cdot\dfrac{-5x+5}{2\cdot(5x-3)}\\\displaystyle\sf f'(1)=\lim_{x \to 1}\dfrac{1}{\diagup\!\!\!\!\!(x-\diagup\!\!\!\!1)}\cdot\dfrac{-5\cdot\diagup\!\!\!\!\!(x-\diagup\!\!\!\!1)}{2\cdot(5x-3)}\\\displaystyle\sf f'(1)=-\dfrac{5}{2}\lim_{x \to 1}\dfrac{1}{5x-3}\\\displaystyle\sf f'(1)=-\dfrac{5}{2}\cdot\dfrac{1}{(5\cdot1-3)}\\\sf f'(1)=-\dfrac{5}{2}\cdot\dfrac{1}{2}$

$\huge\boxed{\boxed{\boxed{\boxed{\sf f'(1)=-\dfrac{5}{4}\checkmark}}}}$