Resolva a derivada:
f(x)= x^2Inx + 2e^x
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Boa tarde Ariely!
Solução!
![f(x)= x^{2}lnx +2e^x\\\\\\
f(x)= x^{2} .lnx+2e^x\\\\\\\
Novamente!\\\\\\\
u= x^{2} ~~~~~v=lnx\\\\\\
u'=2x~~~~~~v'= \dfrac{1}{x}\\\\\\\
Derivada~~de~~2e^x=2e^x](https://tex.z-dn.net/?f=f%28x%29%3D+x%5E%7B2%7Dlnx+%2B2e%5Ex%5C%5C%5C%5C%5C%5C%0Af%28x%29%3D+x%5E%7B2%7D+.lnx%2B2e%5Ex%5C%5C%5C%5C%5C%5C%5C%0ANovamente%21%5C%5C%5C%5C%5C%5C%5C%0Au%3D+x%5E%7B2%7D+%7E%7E%7E%7E%7Ev%3Dlnx%5C%5C%5C%5C%5C%5C%0Au%27%3D2x%7E%7E%7E%7E%7E%7Ev%27%3D+%5Cdfrac%7B1%7D%7Bx%7D%5C%5C%5C%5C%5C%5C%5C%0ADerivada%7E%7Ede%7E%7E2e%5Ex%3D2e%5Ex+ "f(x)= x^{2}lnx +2e^x\\\\\\
f(x)= x^{2} .lnx+2e^x\\\\\\\
Novamente!\\\\\\\
u= x^{2} ~~~~~v=lnx\\\\\\
u'=2x~~~~~~v'= \dfrac{1}{x}\\\\\\\
Derivada~~de~~2e^x=2e^x")
Boa tarde!
Bons estudos!
Solução!
Boa tarde!
Bons estudos!
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