Question

calcule a deriva de 2 tg³+ cos⁴​

Answer 1

A derivada da soma de duas funções é igual a soma das derivadas das funções, logo temos:

$\dfrac{d}{dx}\left(~2tg^3(x)~+~cos^4(x)~\right)~=\\\\\\=~\boxed{\dfrac{d}{dx}\left(~2tg^3(x)~\right)~+~\dfrac{d}{dx}\left(~cos^4(x)~\right)}$

Vamos então calcular as derivadas separadamente e, posteriormente, podemos soma-las.

$\dfrac{d}{dx}\left(~2tg^3(x)~\right)~=\\\\\\Vamos~utilizar~a~regra~da~cadeia\\\\Seja~~\left\{\begin{array}{ccc}f(u(x))&=&2u^3(x)\\u(x)&=&tg(x)\end{array}\right\\\\\\=~\dfrac{d}{du}f(u)~\cdot~\dfrac{d}{dx}u(x)\\\\\\=~2\cdot3\cdot u^{3-1}(x)~~\cdot~~sec^2(x)\\\\\\=~2\cdot3\cdot u^2(x)\cdot sec^2(x)\\\\\\=~6\cdot u^2(x)\cdot sec^2(x)\\\\\\=~\boxed{6\cdot tg^2(x)\cdot sec^2(x)}$

$\dfrac{d}{dx}\left(~cos^4(x)~\right)~=\\\\\\Vamos~utilizar~a~regra~da~cadeia\\\\Seja~~\left\{\begin{array}{ccc}f(u(x))&=&u^4(x)\\u(x)&=&cos(x)\end{array}\right\\\\\\=~\dfrac{d}{du}f(u)~\cdot~\dfrac{d}{dx}u(x)\\\\\\=~4\cdot u^{4-1}(x)~~\cdot~~\left(-sen(x)\right)\\\\\\=\,-4u^3(x)\cdot sen(x)\\\\\\=~\boxed{-4cos^3(x)sen(x)}$

Agora, somando as derivadas calculadas, temos:

$\dfrac{d}{dx}\left(~2tg^3(x)~\right)~+~\dfrac{d}{dx}\left(~cos^4(x)~\right)~=\\\\\\=~\boxed{6tg^2(x)sec^2(x)~-~4cos^3(x)sen(x)}$