Question

Use a regra do quociente para derivar as funções a seguir:
a) f(x) = (3x-2)/(2x+5)
b) f(x) = (x²-1)/(x+1)²

Answer 1

Regra do quociente:

$\boxed{\boxed{\dfrac{d}{dx}\left[\dfrac{f(x)}{g(x)}\right]=\dfrac{f'(x)\cdot g(x)-g'(x)\cdot f(x)}{[g(x)]^{2}}}}$

Regra da cadeia:

$\boxed{\boxed{\dfrac{d}{dx}[f(g(x))]=f'(g(x))\cdot g'(x)}}$
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a)

$f'(x)=\dfrac{[\frac{d}{dx}(3x-2)]\cdot(2x+5)-[\frac{d}{dx}(2x+5)]\cdot(3x-2)}{(2x+5)^{2}}\\\\\\f'(x)=\dfrac{3\cdot(2x+5)-2\cdot(3x-2)}{(2x+5)^{2}}\\\\\\f'(x)=\dfrac{6x+15-6x+4}{(2x+5)^{2}}\\\\\\\boxed{\boxed{f'(x)=\dfrac{19}{(2x+5)^{2}}}}$

b)

$f(x)=\dfrac{x^{2}-1}{(x+1)^{2}}\\\\\\f'(x)=\dfrac{[\frac{d}{dx}(x^{2}-1)]\cdot(x+1)^{2}-[\frac{d}{dx}(x+1)^{2}]\cdot(x^{2}-1)}{[(x+1)^{2}]^{2}}\\\\\\f'(x)=\dfrac{2x\cdot(x+1)^{2}-[\frac{d}{dx}(x+1)^{2}]\cdot(x^{2}-1)}{(x+1)^{4}}$

A derivada de (x + 1)² pode ser calculada pela regra da cadeia:

$\dfrac{d}{dx}(x+1)^{2}=2(x+1)^{2-1}\cdot\dfrac{d}{dx}(x+1)\\\\\\\dfrac{d}{dx}(x+1)^{2}=2(x+1)\cdot1\\\\\\\dfrac{d}{dx}(x+1)^{2}=2(x+1)$

Então:

$f'(x)=\dfrac{2x\cdot(x+1)^{2}-2\cdot(x+1)\cdot(x^{2}-1)}{(x+1)^{4}}$

Colocando (x + 1) em evidência:

$f'(x)=\dfrac{(x+1)\cdot[2x\cdot(x+1)-2\cdot(x^{2}-1)]}{(x+1)^{4}}\\\\\\f'(x)=\dfrac{2x(x+1)-2(x^{2}-1)}{(x+1)^{3}}\\\\\\f'(x)=\dfrac{2x(x+1)-2(x^{2}-1^{2})}{(x+1)^{3}}\\\\\\f'(x)=\dfrac{2x(x+1)-2(x+1)(x-1)}{(x+1)^{3}}\\\\\\f'(x)=\dfrac{(x+1)\cdot[(2x-2(x-1)])}{(x+1)^{3}}\\\\\\f'(x)=\dfrac{2x-2x+2}{(x+1)^{2}}\\\\\\\boxed{\boxed{f'(x)=\dfrac{2}{(x+1)^{2}}}}$