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derivative of 4e^e^x
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4e^(x+e^x)
Chain rule.
\[4e^{e^x}\]
4e^(e^x)e^x
im confused now
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For this problem, you are required to use chain rule, which is: \[F'(x) = f'(g(x))g'(x)\] or \[\frac{ dy }{ dx } = \frac{ dy }{ du }\frac{ du }{ dx }\]
1. factor out the constant (4) 2. take the derivative of e (which is also like f(x) 3. take the derivative of the inside function ( which is like g(x)) where f(x) = e and g(x) = e^x
oh i see now, so basically it stays the same?
Yes, but you have to also multiply it by the g(x) as you can see by the chain rule. So it kinda stays the sme but also looks slightly different.
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