Seja f : R → R derivavel tal que g(1) =2 e g'(1) =3 . Calcule f'(0) sendo que f dada por f(x) = e^x g (3x+1)
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Primeiro, vamos derivar a expressão de f(x):
![<br />f(x)=e^xg(3x+1)\\\\<br />f'(x)=[e^x]'g(3x+1)+e^x[g(3x+1)]'\\\\<br />f'(x)=e^xg(3x+1)+e^xg'(3x+1)\cdot3\\\\<br />f'(x)=e^x[g(3x+1)+3g'(3x+1)](https://tex.z-dn.net/?f=%3Cbr+%2F%3Ef%28x%29%3De%5Exg%283x%2B1%29%5C%5C%5C%5C%3Cbr+%2F%3Ef%27%28x%29%3D%5Be%5Ex%5D%27g%283x%2B1%29%2Be%5Ex%5Bg%283x%2B1%29%5D%27%5C%5C%5C%5C%3Cbr+%2F%3Ef%27%28x%29%3De%5Exg%283x%2B1%29%2Be%5Exg%27%283x%2B1%29%5Ccdot3%5C%5C%5C%5C%3Cbr+%2F%3Ef%27%28x%29%3De%5Ex%5Bg%283x%2B1%29%2B3g%27%283x%2B1%29 "<br />f(x)=e^xg(3x+1)\\\\<br />f'(x)=[e^x]'g(3x+1)+e^x[g(3x+1)]'\\\\<br />f'(x)=e^xg(3x+1)+e^xg'(3x+1)\cdot3\\\\<br />f'(x)=e^x[g(3x+1)+3g'(3x+1)")
Agora, queremos f'(0), então, para x=0 na expressão acima:
![<br />f'(0)=e^0[g(3\cdot0+1)+3g'(3\cdot0+1)]\\\\<br />f'(0)=1\cdot[g(0+1)+3g'(0+1)]\\\\<br />f'(0)=g(1)+3g'(1)\\\\<br />f'(0)=2+3\cdot3\\\\<br />f'(0)=2+9\\\\<br />\boxed{f'(0)=11}](https://tex.z-dn.net/?f=%3Cbr+%2F%3Ef%27%280%29%3De%5E0%5Bg%283%5Ccdot0%2B1%29%2B3g%27%283%5Ccdot0%2B1%29%5D%5C%5C%5C%5C%3Cbr+%2F%3Ef%27%280%29%3D1%5Ccdot%5Bg%280%2B1%29%2B3g%27%280%2B1%29%5D%5C%5C%5C%5C%3Cbr+%2F%3Ef%27%280%29%3Dg%281%29%2B3g%27%281%29%5C%5C%5C%5C%3Cbr+%2F%3Ef%27%280%29%3D2%2B3%5Ccdot3%5C%5C%5C%5C%3Cbr+%2F%3Ef%27%280%29%3D2%2B9%5C%5C%5C%5C%3Cbr+%2F%3E%5Cboxed%7Bf%27%280%29%3D11%7D "<br />f'(0)=e^0[g(3\cdot0+1)+3g'(3\cdot0+1)]\\\\<br />f'(0)=1\cdot[g(0+1)+3g'(0+1)]\\\\<br />f'(0)=g(1)+3g'(1)\\\\<br />f'(0)=2+3\cdot3\\\\<br />f'(0)=2+9\\\\<br />\boxed{f'(0)=11}")
Agora, queremos f'(0), então, para x=0 na expressão acima:
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