Question

Derivada ((x^2-1)cos(2x))/2x

Answer 1

Aplicaremos a regra do produto e a regra do quociente nesta função.

$\sf Regra~do~Produto:~\boxed{\sf \dfrac{d}{dx}\big(f(x)\cdot g(x)\big)~=~\dfrac{d}{dx}f(x)\cdot g(x)+\dfrac{d}{dx}g(x)\cdot f(x)}\\\\\\Regra~do~Quociente:~\boxed{\sf \dfrac{d}{dx}\left(\dfrac{f(x)}{g(x)}\right)~=~\dfrac{\dfrac{d}{dx}f(x)\cdot g(x)-\dfrac{d}{dx}g(x)\cdot f(x)}{g^2(x)}}$

Antes, no entanto, vamos calcular as derivadas de (x²-1), cos(2x) e 2x individualmente para facilitar.

$\sf \dfrac{d(x^2-1)}{dx}~=~\dfrac{d(x^2)}{dx}~-~\dfrac{d(1)}{dx}\\\\\\\dfrac{d(x^2-1)}{dx}~=~2\cdot x^{2-1}~-~0\\\\\\\boxed{\sf \dfrac{d(x^2-1)}{dx}~=~2x}\\\\\\\\\dfrac{d(2x)}{dx}~=~2\cdot \dfrac{d(x)}{dx}\\\\\\\dfrac{d(2x)}{dx}~=~2\cdot 1\cdot x^{1-1}\\\\\\\dfrac{d(2x)}{dx}~=~2\cdot 1\\\\\\\boxed{\sf \dfrac{d(2x)}{dx}~=~2}$

$\sf \dfrac{d(cos(2x))}{dx}~=~?\\\\Aplicar~a~Regra~da~Cadeia:~~ \boxed{\sf \dfrac{d}{dx}f(u)~=~\dfrac{d}{du}f(u)\cdot \dfrac{d}{dx}u(x)}\\\\\\\dfrac{d(cos(2x))}{dx}~=~\dfrac{d}{du}cos(u)\cdot \dfrac{d}{dx}(2x)~~~~~~~Para~u=2x\\\\\\\dfrac{d(cos(2x))}{dx}~=\,-sen(u)\cdot 2\\\\\\\boxed{\sf \dfrac{d(cos(2x))}{dx}~=\,-2sen(2x)}$

Prosseguindo, vamos agora calcular a derivada do produto (x²-1).cos(2x):

$\sf \dfrac{d}{dx}\big((x^2-1)\cdot cos(2x)\big)~=~\dfrac{d}{dx}(x^2-1)\cdot cos(2x)~+~\dfrac{d}{dx}cos(2x)\cdot (x^2-1)\\\\\\\boxed{\sf \dfrac{d}{dx}\big((x^2-1)\cdot cos(2x)\big)~=~2x\cdot cos(2x)~-~2sen(2x)\cdot (x^2-1)}$

Por fim, podemos calcular a derivada do quociente entre o produto (x²-1).cos(2x) e 2x:

$\sf \dfrac{d}{dx}\left(\dfrac{(x^2-1)\cdot cos(2x)}{2x}\right)~=\\\\\\=~\dfrac{\dfrac{d}{dx}\big((x^2-1)\cdot cos(2x)\big)\cdot 2x~-~\dfrac{d}{dx}(2x)\cdot \big((x^2-1)\cdot cos(2x)\big)}{(2x)^2}$

$\sf=~\dfrac{\left(2x\cdot cos(2x)~-~2sen(2x)\cdot (x^2-1)\right)\cdot 2x~-~2\cdot \big((x^2-1)\cdot cos(2x)\big)}{(2x)^2}\\\\\\=~\dfrac{4x^2\cdot cos(2x)~-~4x\cdot sen(2x)\cdot (x^2-1)~-~2x^2cos(2x)+2cos(2x)}{4x^2}\\\\\\=~\dfrac{2x^2\cdot cos(2x)~-~4x\cdot sen(2x)\cdot (x^2-1)~+2cos(2x)}{4x^2}\\\\\\=~\dfrac{cos(2x)\cdot (x^2+1)~-~2x\cdot sen(2x)\cdot (x^2-1)}{2x^2}$

$\sf =~\boxed{\sf \dfrac{-2x^3sen(2x)~+~x^2cos(2x)~+~2x\cdot sen(2x)~+~cos(2x)}{2x^2}}$

$\Huge{\begin{array}{c}\Delta \tt{\!\!\!\!\!\!\,\,o}\!\!\!\!\!\!\!\!\:\,\perp\end{array}}Qualquer~d\acute{u}vida,~deixe~ um~coment\acute{a}rio$