Question

Derive $y = x^{2} . sen(x) . tg(x)$

Answer 1

Resposta:

y =x²*sen(x) * tan(x)

y =x²*sen(x) * sen(x)/cos(x)

y =x²*sen²(x) /cos(x)

y'=[(x²*sen²(x))'*cos(x) - x²*sen²(x) * (cos(x))']/cos²(x)

y'=[(2x*sen²(x)+x²*2*sen(x)*cos(x))*cos(x) - x²*sen²(x) * (-sen(x))]/cos²(x)

Answer 2

Derivar a função $\mathtt{y=x^2\,sen(x)\,tg(x).}$

Vamos aplicar a regra para a derivada do produto:

    $\mathtt{\Longrightarrow\quad \dfrac{dy}{dx}=\dfrac{d}{dx}\big[x^2\,sen(x)\,tg(x)\big]}\\\\\\\\\mathtt{\Longleftrightarrow\quad \dfrac{dy}{dx}=\dfrac{d}{dx}\big[x^2\,sen(x)\big]\cdot tg(x)+x^2\,sen(x)\cdot \dfrac{d}{dx}\big[tg(x)\big]}$

Derive x² sen(x) aplicando a regra do produto novamente:

    $\mathtt{\Longleftrightarrow\quad \dfrac{dy}{dx}=\left[\dfrac{d}{dx}(x^2)\cdot sen(x)+x^2\cdot \dfrac{d}{dx}\big[sen(x)\big]\right]\cdot tg(x)+x^2\,sen(x)\cdot \dfrac{d}{dx}\big[tg(x)\big]}\\\\\\\\\mathtt{\Longleftrightarrow\quad \dfrac{dy}{dx}=\big[2x\,sen(x)+x^2\,cos(x)\big]\cdot tg(x)+x^2\,sen(x)\cdot \dfrac{d}{dx}\big[tg(x)\big]}$

Por fim, derivando tg(x) obtemos sec²(x):

    $\mathtt{\Longleftrightarrow\quad \dfrac{dy}{dx}=\big[2x\,sen(x)+x^2\,cos(x)\big]\cdot tg(x)+x^2\,sen(x)\,sec^2(x)}$

    $\mathtt{\Longleftrightarrow\quad \dfrac{dy}{dx}=2x\,sen(x)\,tg(x)+x^2\,cos(x)\,tg(x)+x^2\,sen(x)\,sec^2(x)\quad\longleftarrow\quad resposta.}$

Dúvidas? Comente.

Bons estudos! :-)