Question

Calcular a função derivada de cada uma das funções a seguir

Answer 1

Resposta:

Explicação passo-a-passo:

Lembre da Regra do Quociente:

[u/v]' = [u'v - v'u]/v²

1)

y' = [x / 1 + 2x]' = [1 (1+2x) - x(2)] / (1 + 2x)²

y' = [1 + 2x - 2x] / (1 + 2x)²

y' = [1] / (1 + 2x)²

2)

y' = [3 / x + 1]' = [0 (x + 1) - 3 (1)] / (x + 1)²

y' = [0 - 3]  / (x + 1)²

y' = -3  / (x + 1)²

3)

y' = [2x / x + 10]' = [2 (x + 10) - 2x (1)] / (x + 10)²

y' = [2x + 20 - 2x] / (x + 10)²

y = 20 / (x + 10)²

4)

y' = [(x +1) / (x - 1)]' = [(1) (x - 1) - {(x + 1) (1)}] / (x - 1)²

y' = [x - 1 - x - 1] / (x - 1)²

y' = [-1 -1] / (x - 1)²

y' = -2 / (x - 1)²

Answer 2

Explicação passo-a-passo:

Cálculo das derivadas de funções:

$\mathsf{A)~y=\dfrac{x}{1+2x}' }$

Regra do Quociente ( Aplicação ) :

$\mathsf{\dfrac{f(x)}{g(x)}=\dfrac{f(x)'.g(x)-f(x).g(x)'}{\Big(g(x)\Big)^2} }$

$\mathsf{y'=\dfrac{x'.(1+2x)-x.(1+2x)'}{\Big(1+2x\Big)^2} }$

$\mathsf{y'=\dfrac{1.(1+2x)-x.2}{(1+2x)^2 } }$

$\mathsf{y'=\dfrac{1+2x-2x}{(1+2x)^2} }$

$\mathsf{y'=\dfrac{1}{(1+2x)^2} }$

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$\mathsf{B)~y=\dfrac{3}{x+1} }$

$\mathsf{y'=\dfrac{3'(x+1)-3.(x+1)'}{(x+1)^2} }$

$\mathsf{y'=-\dfrac{3}{(x+1)^2} }$

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$\mathsf{C)~y=\dfrac{2x}{x+10}' }$

$\mathsf{y'=\dfrac{(2x)'(x+10)-2x.(x+10)' }{(x+10)^2} }$

$\mathsf{y'=\dfrac{2.(x+10)-2x}{(x+10)^2} }$

$\mathsf{y'=\dfrac{2x+20-2x}{(x+10)^2} }$

$\mathsf{y'=\dfrac{20}{(x+10)^2} }$

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$\mathsf{D)~y=\dfrac{x+1}{x-1} }$

$\mathsf{y'=\dfrac{(x+1)'(x-1)-(x+1).(x-1)'}{(x-1)^2} }$

$\mathsf{y'=\dfrac{(x-1)-(x+1)}{(x-1)^2} }$

$\mathsf{y'=\dfrac{x-1-x-1}{(x-1)^2} }$

$\mathsf{y'=-\dfrac{2}{(x-1)^2} }$

Espero ter ajudado bastante!)