Question

Calcule as derivadas abaixo.

3) y = (x³ + 1).senx

$4)y = \sqrt[3]{ ln( {x}^{5} + senx ) }$
passo a passo.
grato. ​

Answer 1

Explicação passo-a-passo:

3) $\sf y=(x^3+1)\cdot\text{sen}~(x)$

$\sf \dfrac{dy}{dx}=(x^3+1)'\cdot\text{sen}~(x)+[\text{sen}(x)]'\cdot(x^3+1)$

$\sf \dfrac{dy}{dx}=3x^2\cdot\text{sen}~(x)+[\text{cos}~(x)]\cdot(x^3+1)$

$\sf \dfrac{dy}{dx}=(x^3+1)\cdot\text{cos}~(x)+3x^2\cdot\text{sen}~(x)$

4) $\sf y=\sqrt[3]{\text{ln}~(x^5+\text{sen}~x)}$

$\sf y=[\text{ln}~(x^5+\text{sen}~x)]^{\frac{1}{3}}$

$\sf \dfrac{dy}{dx}=[\text{ln}~(x^5+\text{sen}~x)]'\cdot\dfrac{1}{3}\cdot[\text{ln}~(x^5+\text{sen}~x)]^{\frac{1}{3}-1}$

$\sf \dfrac{dy}{dx}=\dfrac{(x^5+\text{sen}~x)'}{x^5+\text{sen}~x}\cdot\dfrac{1}{3}\cdot[\text{ln}~(x^5+\text{sen}~x)]^{\frac{-2}{3}}$

$\sf \dfrac{dy}{dx}=\dfrac{5x^4+\text{cos}~x}{x^5+\text{sen}~x}\cdot\dfrac{1}{3}\cdot\dfrac{1}{\sqrt[3]{[\text{ln}~(x^5+\text{sen}~x)]^2}}$

$\sf \dfrac{dy}{dx}=\dfrac{5x^4+\text{cos}~x}{3\cdot(x^5+\text{sen}~x)\cdot\sqrt[3]{[\text{ln}~(x^5+\text{sen}~x)]^2}}$

Answer 2

• A)

$\sf y\:=\:\left(x^3\:+\:1\right)\cdot \:senx\\\\\\\\\sf \displaystyle \frac{d}{dx}\left(\left(x^3+1\right)sin \left(x\right)\right)\\\\\\\\\sf Lembre \: de \:{Aplicar\:a\:regra\:do\:produto}:\quad \left(f\cdot g\right)'=f\:'\cdot g+f\cdot g'\\\\\\\\\sf \frac{d}{dx}\left(x^3+1\right)sin \left(x\right)+\frac{d}{dx}\left(sin \left(x\right)\right)\left(x^3+1\right)\\\\\\\\ \frac{d}{dx}\left(x^3+1\right)=3x^2\\\\\\\\\sf \frac{d}{dx}\left(sin \left(x\right)\right)=cos(x)$

$\boxed{\boxed{\sf 3x^2sin \left(x\right)+cos \left(x\right)\left(x^3+1\right)}}$

• B)

$\sf \displasytyle \large{\sf y\:=\:\sqrt[\sf 3]{\sf in\left(x^5+sen\:x\right)}}\\\\\\\\\sf Aplique \: a \: Regra \: de \: cadeia \Rightarrow \dfrac{1}{3\left(in\left(x^5+sin \left(x\right)\right)\right)^{\frac{2}{3}}}\frac{d}{dx}\left(in\left(x^5+sin \left(x\right)\right)\right)\\\\\\\\\sf \displaystyle \frac{1}{3\left(in\left(x^5+sin \left(x\right)\right)\right)^{\frac{2}{3}}}\frac{d}{dx}\left(in\left(x^5+sin \left(x\right)\right)\right)\\\\\\\\\sf$

$\sf \displaystyle \frac{d}{dx}\left(in\left(x^5+sin \left(x\right)\right)\right)=\frac{1}{3\left(in\left(x^5+sin \left(x\right)\right)\right)^{\frac{2}{3}}}i\left(ncos \left(x\right)+5nx^4\right)\\\\\\\\\boxed{\boxed{\sf \frac{1}{3\left(in\left(x^5+sin \left(x\right)\right)\right)^{\frac{2}{3}}}i\left(ncos \left(x\right)+5nx^4\right)}}$