Question

Calcule as derivadas:
a) y= $(x^2 + 3x/x+1)^3$

b) y= $5^x x^5 + 5e^x - \sqrt{5}$

Answer 1

Usaremos as regras do produto, quociente e da cadeia:

$\boxed{\boxed{\big[f(x)g(x)\big]'=f'(x)g(x)+f(x)g'(x)}}~~\mathtt{(Regra~do~produto)}\\\\\\\boxed{\boxed{\bigg[\dfrac{f(x)}{g(x)}\bigg]'=\dfrac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^{2}}}}~~~\mathtt{(Regra~do~quociente)}\\\\\\\boxed{\boxed{\big[f(g(x))\big]'=f'(g(x))\cdot g'(x)}}~~~\mathtt{(Regra~da~cadeia)}$

Notação: $\log(x)=log_{e}(x)=ln(x)$
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a)

$y=\bigg(\dfrac{x^{2}+3x}{x+1}\bigg)^{3}$

Note que, se fizermos $f(x)=x^{3}$ e $g(x)=\dfrac{x^{2}+3x}{x+1}$, então $y=f(g(x))$.

Portanto, pela regra da cadeia:

$y'=f'(g(x))g'(x)=f'\big(\frac{x^{2}+3x}{x+1}\big)g'(x)$

Encontrando as derivadas de $f$ e $g$:

Derivamos $f$ pela regra da potência:

$f'(x)=\big(x^{3}\big)'=3x^{2}$

Derivamos $g$ pela regra do quociente (junta com a da potência)

$g'(x)=\bigg(\dfrac{x^{2}+3x}{x+1}\bigg)'=\dfrac{\big(x^{2}+3x\big)'\cdot(x+1)-(x^{2}+3x)\cdot\big(x+1\big)'}{(x+1)^{2}}\\\\\\g'(x)=\dfrac{(2x+3)(x+1)-(x^{2}+3x)\cdot1}{(x+1)^{2}}\\\\\\g'(x)=\dfrac{2x^{2}+2x+3x+3-(x^{2}+3x)}{(x+1)^{2}}\\\\\\\boxed{\boxed{g'(x)=\dfrac{x^{2}+2x+3}{(x+1)^{2}}}}$

Portanto, temos

$y'=f'(g(x))\cdot g'(x)\\\\y'=3\big(g(x)\big)^{2}\cdot\dfrac{x^{2}+2x+3}{(x+1)^{2}}\\\\\\y'=3\dfrac{(x^{2}+3x)^{2}}{(x+1)^{2}}\cdot\dfrac{x^{2}+2x+3}{(x+1)^{2}}\\\\\\\boxed{\boxed{y'=\dfrac{3(x^{2}+3x)^{2}(x^{2}+2x+3)}{(x+1)^{4}}}}$
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b)

$y=5^{x}x^{5}+5e^{x}-\sqrt{5}$

Sabemos que $\big(e^{x}\big)'=e^{x}$

mas $5^{x}=e^{\log(5^{x})}=e^{x\log(5)}$

Derivando pela regra da cadeia, obtemos

$\big(5^{x}\big)'=\big(e^{x\log(5)}\big)'=e^{x\log(5)}\cdot\log(5)=5^{x}\log(5)$

(Generalizando, temos $\big(a^{x}\big)'=a^{x}\log(a)$ para $a~\textgreater~0$)

Além disso, note que $\big(cf(x)\big)'=\big(c\big)'f(x)+cf'(x)=0\cdot f(x)+cf'(x)=cf'(x)$, onde $c\in\mathbb{R}$

Então, como a derivada da soma é soma das derivadas:

$y'=\big(5^{x}x^{5}+5e^{x}-\sqrt{5})'=(5^{x}x^{5})'+(5e^{x})'+(-\sqrt{5})'\\\\y'=\big[(5^{x})'x^{5}+5^{x}(x^{5})'\big]+5(e^{x})'+0\\\\y'=\big[5^{x}\log(5)x^{5}+5^{x}\cdot 5x^{4}\big]+5e^{x}\\\\\boxed{\boxed{y'=5^{x}x^{5}\log(5)+5^{x+1}x^{4}+5e^{x}}}$