Question

Qual é a derivada de:
h(X)= x^4/4 ln x - x^4/16

Answer 1

Calcular a derivada da função

$\mathsf{h(x)=\dfrac{x^4}{4}\,\ell n\,x-\dfrac{x^4}{16}}$

Aplicando as regras de derivação:

$\mathsf{\dfrac{d}{dx}\big[h(x)\big]=\dfrac{d}{dx}\Big(\dfrac{x^4}{4}\,\ell n\,x-\dfrac{x^4}{16}\Big)}\\\\\\ \mathsf{\dfrac{d}{dx}\big[h(x)\big]=\dfrac{d}{dx}\Big(\dfrac{x^4}{4}\,\ell n\,x\Big)-\dfrac{d}{dx}\Big(\dfrac{x^4}{16}\Big)}\\\\\\ \mathsf{\dfrac{d}{dx}\big[h(x)\big]=\bigg[\dfrac{d}{dx}\Big(\dfrac{x^4}{4}\Big)\cdot \ell n\,x+\dfrac{x^4}{4}\cdot \dfrac{d}{dx}(\ell n\,x)\bigg]-\dfrac{d}{dx}\Big(\dfrac{x^4}{16}\Big)}\\\\\\ \mathsf{\dfrac{d}{dx}\big[h(x)\big]=\bigg[\dfrac{1}{4}\cdot 4x^{4-1}\cdot \ell n\,x+\dfrac{x^4}{4}\cdot \dfrac{1}{x}\bigg]-\dfrac{1}{16}\cdot 4x^{4-1}}\\\\\\ \mathsf{\dfrac{d}{dx}\big[h(x)\big]=\bigg[x^3\,\ell n\,x+\dfrac{x^3}{4}\bigg]-\dfrac{x^3}{4} }$

$\mathsf{\dfrac{d}{dx}\big[h(x)\big]=x^3\,\ell n\,x \quad\longleftarrow\quad resposta.}$

Dúvidas? Comente.

Bons estudos! :-)