Question

Derive as funções a seguir:

Answer 1

$\large\boxed{\begin{array}{l}\underline{\sf Regras\,b\acute asicas\,de\,derivac_{\!\!,}\tilde ao}\\\rm 1)~\dfrac{d}{dx}(k)=0,sendo\,k\,constante\\\\\rm 2)~\dfrac{d}{dx}(f(x)\pm g(x))=\dfrac{d}{dx}f(x)\pm\dfrac{d}{dx}g(x)\\\\\rm 3)~\dfrac{d}{dx}(u^n)=nu^{n-1}\cdot\dfrac{du}{dx}\\\\\rm 4)~\dfrac{d}{dx}(f(x)\cdot g(x))=\dfrac{d}{dx}f(x)\cdot g(x)+f(x)\cdot\dfrac{d}{dx}g(x)\\\\\rm 5)~\dfrac{d}{dx}\bigg(\dfrac{f(x)}{g(x)}\bigg)=\dfrac{\dfrac{d}{dx}f(x)\cdot g(x)-f(x)\cdot\dfrac{d}{dx}g(x)}{g(x)^2}\end{array}}$

$\large\boxed{\begin{array}{l}\underline{\sf Derivada\,de\,func_{\!\!,}\tilde oes\,trigonom\acute etricas}\\\rm 1)~\dfrac{d}{dx}[sen(u)]=cos(u)\cdot\dfrac{du}{dx}\\\\\rm 2)~\dfrac{d}{dx}[cos(u)]=-sen(u)\cdot\dfrac{du}{dx}\\\\\rm 3)~\dfrac{d}{dx}tg(u)=sec^2(u)\cdot\dfrac{du}{dx}\\\\\rm 4)~\dfrac{d}{dx}[sec(u)]=sec(u)\cdot tg(u)\cdot\dfrac{du}{dx}\\\\\rm 5)~\dfrac{d}{dx}[cossec(u)]=-cossec(u)\cdot cotg(u)\cdot\dfrac{du}{dx}\\\\\rm 6)~\dfrac{d}{dx}[cotg(u)]=-cosec^2(u)\cdot\dfrac{du}{dx}\end{array}}$

$\large\boxed{\begin{array}{l}\underline{\sf Derivada\,de\,func_{\!\!,}\tilde oes\,exponenciais}\\\rm 1)~quando\,a\,base\,\acute e~\bf{e}\\\rm\dfrac{d}{dx}[e^u]=e^u\cdot\dfrac{du}{dx}\\\\\rm 2)~Quando\,se\,trata\,de\,base\,qualquer\\\rm\dfrac{d}{dx}[a^u]=a^u\cdot \ln a\cdot\dfrac{du}{dx}\end{array}}$

$\large\boxed{\begin{array}{l}\underline{\sf Derivada\,de\,func_{\!\!,}\tilde oes\,logar\acute itmicas}\\\rm 1)~quando\,a\,base\,\acute e~\bf e\\\rm\dfrac{d}{dx}[\ln u]=\dfrac{1}{u}\cdot\dfrac{du}{dx}\\\\\rm 2)~\dfrac{d}{dx}[\log_bu]=\dfrac{1}{u\ln b}\cdot\dfrac{du}{dx}\end{array}}$

$\large\boxed{\begin{array}{l}\rm a)~ y(x)=(x^5-3x)^2\\\rm y'(x)=2(x^5-3x)^{2-1}\cdot(5x^4-3)\\\rm y'(x)=(10x^4-6)(x^5-3x)\end{array}}$

$\large\boxed{\begin{array}{l}\rm b)~ y(x)=\sqrt{x^2-3}\\\rm y(x)=(x^2-3)^{\frac{1}{2}}\\\rm y'(x)=\dfrac{1}{\diagup\!\!\!2}(x^2-3)^{\frac{1}{2}-1}\cdot(\diagup\!\!\!2x)\\\\\rm y'(x)=\dfrac{x}{(x^2-3)^{\frac{1}{2}}}=\dfrac{x}{\sqrt{x^2-3}}\end{array}}$

$\large\boxed{\begin{array}{l}\rm c)~ y(x)=sen(x)-cos(x)\\\rm y'(x)=cos(x)-(-sen(x))\\\rm y'(x)=cos(x)+sen(x)\end{array}}$

$\large\boxed{\begin{array}{l}\rm d)~ y(x)=(sen(x)-cos(x))^3\\\rm y'(x)=3(sen(x)-cos(x))^{3-1}\cdot(cos(x)-(-sen(x)))\\\rm y'(x)=3(sen(x)-cos(x))^2(cos(x)+sen(x))\end{array}}$

$\large\boxed{\begin{array}{l}\rm e)~y(x)=x^4\cdot e^x\\\rm y'(x)=4x^3\cdot e^x+x^4\cdot e^x\\\rm coloca\,x^3\,e^x\,em\,evid\hat encia\\\rm y'(x)=x^3e^x(4+x)\end{array}}$

$\large\boxed{\begin{array}{l}\rm f)~ y(x)=\dfrac{e^x}{x^2}\\\\\rm y'(x)=\dfrac{e^x\cdot x^2-e^x\cdot 2x}{(x^2)^2}\\\\\\\rm y'(x)=\dfrac{x^2e^x-2x\,e^x}{x^4}\\\rm coloca\,xe^x\,em\,evid\hat encia\\\rm y'(x)=\dfrac{\diagup \!\!\!\!x\cdot e^x(x-2)}{\diagup\!\!\!\!\!x^4}\\\\\rm y'(x)=\dfrac{e^x(x-2)}{x^3}\end{array}}$

$\large\boxed{\begin{array}{l}\rm g)~ y(x)=\sqrt{e^x+x}\\\rm y(x)=(e^x+x)^{\frac{1}{2}}\\\rm y'(x)=\dfrac{1}{2}(e^x+x)^{\frac{1}{2}-1}\cdot(e^x+1)\\\\\rm y'(x)=\dfrac{e^x+1}{2\sqrt{e^x+x}}\end{array}}$

$\large\boxed{\begin{array}{l}\rm h)~ y(x)=x^2+\sqrt{x}+x^{\frac{1}{3}}\\\rm y(x)=x^2+x^{\frac{1}{2}}+x^{\frac{1}{3}}\\\rm y'(x)=2x+\dfrac{1}{2}x^{\frac{1}{2}-1}+\dfrac{1}{3}x^{\frac{1}{3}-1}\\\\\rm y'(x)=2x+\dfrac{1}{2}x^{-\frac{1}{2}}+\dfrac{1}{3}x^{-\frac{2}{3}}\\\\\rm y'(x)=2x+\dfrac{1}{2\sqrt{x}}+\dfrac{1}{3\sqrt[\rm3]{\rm x^2}}\end{array}}$

$\large\boxed{\begin{array}{l}\rm i)~ y(x)= sen^3x+cos^2(x)\\\rm y'(x)=3sen^2(x)\cdot cos(x)+2cos(x)\cdot[-sen(x)]\\\rm y'(x)=3sen^2(x)\cdot cos(x)-2sen(x)\cdot cos(x)\\\rm coloca\,sen(x)\cdot cos(x)\,em\,evid\hat encia\\\rm y'(x)=sen(x)\cdot cos(x)[3sen(x)-2]\end{array}}$

$\large\boxed{\begin{array}{l}\rm j)~y(x)= 2^{x^2+3x}\\\rm y'(x)=2^{x^2+3x}\cdot\ln 2\cdot(2x+3)\end{array}}$

Answer 2

Resposta:

y' = 2x(5x^8 - 18x^4 - 9)

Explicação passo a passo:

y = (x^5 - 3x)²

y' = 2(x^5 - 3x)(5x^4 - 3)

y' = 2(5x^9 - 3x^5 - 15x^5 - 9x)

y' = 2(5x^9 - 18x^5 - 9x)

y' = 2x(5x^8 - 18x^4 - 9)