Question

Derive as funções a seguir:

Answer 1

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

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

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

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

$\large\boxed{\begin{array}{l}\sf 1)~ y(x)=x^3\cdot e^x\\\rm y'(x)=3x^{3-1}\cdot e^x+x^3\cdot e^x\\\rm y'(x)=3x^2e^x+x^3e^x\\\rm coloca\,x^2e^x~em~evid\hat encia\\\rm y'(x)=x^2e^x(3+x)\end{array}}$

$\large\boxed{\begin{array}{l}\sf 2)~\rm f(t)=\dfrac{e^t}{t^2}\\\\\rm f'(t)=\dfrac{e^t\cdot t^2-e^t\cdot 2t}{(t^2)^2}\\\\\rm f'(t)=\dfrac{t^2e^t-2te^t}{t^4}\\\rm coloca\,te^t\,em\,evid\hat encia:\\\rm f'(t)=\dfrac{\backslash\!\!\!t\cdot e^t(t-2))}{\diagup\!\!\!\!\!t^4}\\\\\rm f'(t)=\dfrac{e^t(t-2)}{t^3}\end{array}}$

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

$\large\boxed{\begin{array}{l}\sf 4)~\rm y=2^{t^2+t}\\\rm \dfrac{dy}{dt}=2^{t^2+t}\cdot \ln 2\cdot(2t+1)\end{array}}$

$\large\boxed{\begin{array}{l}\sf 5)~\rm y=x^5+\sqrt{x}+\dfrac{1}{x}\\\rm vamos\,escrever\,a\,func_{\!\!,}\tilde ao\,de\,modo\,conveniente\\\rm y=x^5+x^{\frac{1}{2}}+x^{-1}\\\rm \dfrac{dy}{dx}=5x^{5-1}+\dfrac{1}{2}x^{\frac{1}{\!2}-1}-1x^{-1-1}\\\\\rm\dfrac{dy}{dx}=5x^4+\dfrac{1}{2}x^{-\frac{1}{2}}-x^{-2}\\\\\rm\dfrac{dy}{dx}=5x^4+\dfrac{1}{2\sqrt{x}}-\dfrac{2}{x^2}\end{array}}$

$\large\boxed{\begin{array}{l}\sf 6)~\rm y=sen^2(x)\\\rm \dfrac{dy}{dx}=2sen(x)\cdot cos(x)\\\\\rm\dfrac{dy}{dx}=sen(2x)\end{array}}$

$\large\boxed{\begin{array}{l}\sf 7)~\rm y=\sqrt{x^5+3x^2}\\\rm y=(x^5+3x^2)^{\frac{1}{2}}\\\rm\dfrac{dy}{dx}=\dfrac{1}{2}(x^5+3x^2)^{-\frac{1}{2}}\cdot(5x^4+6x)\\\\\rm\dfrac{dy}{dx}=\dfrac{5x^4+6x}{2\sqrt{x^5+3x^2}}\end{array}}$

$\large\boxed{\begin{array}{l}\sf 8)~\rm y=\dfrac{x^2+x}{e^x}\\\\\rm\dfrac{dy}{dx}=\dfrac{(2x+1)\cdot e^x-(x^2+x)\cdot e^x}{(e^x)^2}\\\\\rm\dfrac{dy}{dx}=\dfrac{\diagup\!\!\!\!e^x(2x+1-x^2-x)}{\diagup\!\!\!\!e^x\cdot e^x}\\\\\rm\dfrac{dy}{dx}=\dfrac{-x^2+x+1}{e^x}\\\\\rm\dfrac{dy}{dx}=-\dfrac{x^2-x-1}{e^x}\end{array}}$

$\large\boxed{\begin{array}{l}\sf 9)~\rm y=\ln(x^2+3x^5)\\\rm\dfrac{dy}{dx}=\dfrac{1}{x^2+3x^5}\cdot(2x+15x^4)\\\\\rm\dfrac{dy}{dx}=\dfrac{\diagup\!\!\!x(2+15x^3)}{\diagup\!\!\!\!\!x^2(1+3x^3)}\\\\\rm\dfrac{dy}{dx}=\dfrac{2+15x^3}{x(1+3x^3)}\end{array}}$

$\large\boxed{\begin{array}{l}\sf 10)~\rm y=\log_5(x^3+2x)\\\rm\dfrac{dy}{dx}=\dfrac{1}{(x^3+2x)\ln5}\cdot(3x^2+2)\\\\\rm\dfrac{dy}{dx}=\dfrac{3x^2+2}{(x^3+2x)\ln5}\end{array}}$

$\large\boxed{\begin{array}{l}\sf 11)~\rm y=3^{x+sen(x)}\\\rm\dfrac{dy}{dx}=3^{x+sen(x)}\cdot\ln(3)\cdot(1+cos(x))\end{array}}$

$\large\boxed{\begin{array}{l}\sf 12)~\rm y=\dfrac{e^x+x^2}{x^3}\\\\\rm\dfrac{dy}{dx}=\dfrac{(e^x+2x)\cdot x^3-(e^x+x^2)\cdot 3x^2}{(x^3)^2}\\\\\rm\dfrac{dy}{dx}=\dfrac{x^3e^x+2x^4-3x^2e^x-x^4}{x^6}\\\\\rm\dfrac{dy}{dx}=\dfrac{x^3e^x-3x^2e^x+x^4}{x^6}\\\\\rm\dfrac{dy}{dx}=\dfrac{\diagup\!\!\!\!\!x^2(xe^x-3e^x+x^2)}{\diagup\!\!\!\!\!x^6}\\\\\rm\dfrac{dy}{dx}=\dfrac{xe^x-3e^x+x^2}{x^4}\end{array}}$