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What is the fourth derivative of \(f(x)=3xe^x-e^{2x}\) ?
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well f'(x*e^x) = x*e^x + e^x = e^x(x+1). So: f'(x) = 3e^x(x+1) - 2e^(2x) f''(x) = (3xe^x+3e^x - 2e^(2x))' = 3e^x(x+1)+3e^x - 4e^(2x)
f'''(x) = (3xe^x + 3e^x + 3e^x - 4e^(2x))' = (3xe^x + 6e^x - 4e^(2x))' = 3e^x(x+1) + 6e^x - 8e^(2x)
I think I made a mistake in f''(x)
Yes, you did
No i didn't
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You made a mistake somewhere
f''''(x) = (3xe^x+3e^x+6e^x-8e^(2x))' = (3xe^x + 9e^x - 8e^(2x))' = 3e^x(x+1)+9e^x - 16e^(2x) <-Final answer
Btw u didnt go wrong
Thankkksss Bahhhrrrooommmmmmm
Hero.. like i said.. ignore me.
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Yeah, that's the correct answer
Why do you want me to ignore you @bahrom7893
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