How do you differentiate y=e^(e^x)?

1 Answer
Steve M
Nov 5, 2016

dy/dx=e^(x+e^x)

Explanation:

If you are studying maths, then you should learn the Chain Rule for Differentiation, and practice how to use it:

If y=f(x) then f'(x)=dy/dx=dy/(du)(du)/dx

I was taught to remember that the differential can be treated like a fraction and that the "dx's" of a common variable will "cancel" (It is important to realise that dy/dx isn't a fraction but an operator that acts on a function, there is no such thing as "dx" or "dy" on its own!). The chain rule can also be expanded to further variables that "cancel", E.g.

dy/dx = dy/(dv)(dv)/(du)(du)/dx etc, or (dy/dx = dy/color(red)cancel(dv)color(red)cancel(dv)/color(blue)cancel(du)color(blue)cancel(du)/dx)

So we have y=e^(e^x), Then:

{ ("Let "u=e^x, => , (du)/dx=e^x), ("Then "y=e^u, =>, dy/(du)=e^u ) :}

Using dy/dx=(dy/(du))((du)/dx) we get:

dy/dx=(e^u)(e^x)
:. dy/dx=(e^(e^x))(e^x)
:. dy/dx=e^(x+e^x)

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