Question

Resolva as seguintes derivadas com a regra do quociente.

1)

$\boxed{y=\dfrac{x^3-4x}{x^2+1}}$

2)

$\boxed{w=\dfrac{t^2-2t}{3t + 4}}$

Answer 1

Olá !

Primeiro temos que considerar como base a apresentação:

$\boxed{\boxed{(\dfrac{f}{g})' =\dfrac{g.f'-f.g'}{g^{2}} }}$

Agora podemos resolver nossas atividades!

Primeiro:

$y=\dfrac{x^{3}-4x}{x^{2}+1}$

Uma boa, é derivar cada parte antes de começar...

f(x) = x³ - 4x

f(x)' = 3.x³⁻¹ - 1.4x¹⁻¹

f(x)' = 3x² - 4

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g(x) = x² + 1

g(x)' = 2.x²⁻¹ + 0

g(x)' = 2x  

Agora basta aplicar ....

$y'=\dfrac{(x^{2}+1).(3x^{2}-4)-(x^{3}-4x).(2x)}{(x^{2}+1)^{2}} \\\\\\y'=\dfrac{(3x^{4}-4x^{2}+3x^{2}-4)-(2x^{4}-8x^{2})}{x^{4}+2x^{2}+1} \\\\\\y'=\dfrac{3x^{4}-x^{2}-4-2x^{4}+8x^{2}}{x^4+2x^{2}+1} \\\\\\\boxed{\boxed{y'=\dfrac{x^{4}+7x^2-4}{x^{4}+2x^{2}+1} }}$

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Segundo:

Mesma maneira !

f(t) = t² - 2t

f(t)' = 2t²⁻¹ - 1.2t¹⁻¹

f(t)' = 2t - 2

---------------------------------------------------

g(t) = 3t + 4

g(t)' = 1.3t¹⁻¹ + 0.4

g(t)' = 3  

Agora basta aplicar ...

$w=\dfrac{t^{2}-2t}{3t+4} \\\\\\w'=\dfrac{(3t+4).(2t-2)-(t^{2}-2t).(3)}{(3t+4)^{2}} \\\\\\w'=\dfrac{(6t^{2}-6t+8t-8)-(3t^{2}-6t)}{9t^{2}+24t+16} \\\\\\w'=\dfrac{6t^{2}+2t-8-3t^{2}+6t}{9t^{2}+24t+16} \\\\\\\boxed{\boxed{w'=\dfrac{3t^{2}+8t-8}{9t^{2}+24t+16} }}$

Bons estudos!

ok

Answer 2

$\mathsf{y=\dfrac{{x}^{3}-4x}{{x}^{2}+1}}$

$\mathsf{\frac{dy}{dx}=\frac{\frac{d}{dx}({x}^{3}-4x).({x}^{2}+1)-({x}^{3}-4x).\frac{d}{dx}({x}^{2}+1)}{{({x}^{2}+1)}^{2}}}$

$\mathsf{\dfrac{dy}{dx}=\frac{(3{x}^{2}-4)({x}^{2}+1)-({x}^{3}-4x).2x}{{({x}^{2}+1)}^{2}}}$

$\mathsf{ \frac{dy}{dx} = \frac{3{x}^{4}+3{x}^{2}-4{x}^{2} - 4-2{x}^{4}+8{x}^{2}}{{({x}^{2}+1)}^{2}}}$

$\large\boxed{\boxed{\mathsf{\dfrac{dy}{dx}=\dfrac{{x}^{4}+7{x}^{2} - 4}{{({x}^{2}+1)}^{2}}}}}$

b)

$\mathsf{w=\dfrac{t^2-2t}{3t + 4}}$

$\mathsf{\dfrac{dw}{dt}=\frac{\frac{d}{dt}({t}^{2}-2t).(3t+4)-({t}^{2}-2t).\frac{d}{dt}(3t+4)}{{(3t+4)}^{2}}}$

$\mathsf{\dfrac{dw}{dt}=\frac{(2t-2)(3t+4)-({t}^{2}-2t).3}{{(3t+4)}^{2}}}$

$\mathsf{\dfrac{dw}{dt}=\dfrac{6{t}^{2}+2t-8-3{t}^{2}+6t}{{(3t+4)}^{2}}}$

$\large\boxed{\boxed{\mathsf{ \dfrac{dw}{dt}=\dfrac{3{t}^{2}+8t-8}{{(3t+4)}^{2}}}}}$