Question

Derivada de
$f(x)=5x+sen^2(4x)$

Answer 1

$\\ f(x) = 5x +sen^2(4x) = 5x + [sen(4x)]^2 \\ \\ F(x)' = (5x)' + [sen(4x)]^2'*(sen4x)'*(4x)' \\ \\ F(x)' = 5 +2*[sen(4x)]^2^-^1cos(4x)*4 \\ \\ F(x)' = 5 + 8sen(4x)Cos(4x)$

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Propriedade trigonómetrica:

$\\ sen(8x) = 2sen(4x)Cos(4x) \\ \\ \frac{Sen(8x)}{2} = Sen(4x)Cos(4x)$

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$\\ F(x)' = 5 +8* \frac{Sen(8x)}{2} \\ \\ F(x)' = 5 + 4*Sen(8x)$

Answer 2

$\sf \displaystyle f\left(x\right)\:=5x+sen^2\left(4x\right)\\\\\\\frac{d}{dx}\left(5x+\sin ^2\left(4x\right)\right)\\\\\\\:regra\:da\:soma/diferença}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\\\=\frac{d}{dx}\left(5x\right)+\frac{d}{dx}\left(\sin ^2\left(4x\right)\right)\\\\\\\to \boxed{\sf =5+4\sin \left(8x\right)}$