How do you differentiate #y=e^(x+1)+1#?

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Answer 1 · 2017-07-10T17:15:08

Given: #y=e^(x+1)+1#

Differentiate each term:

#dy/dx = (d(e^(x+1)))/dx + (d(1))/dx#

The derivative of a constant is 0:

#dy/dx = (d(e^(x+1)))/dx + 0#

#dy/dx = (d(e^(x+1)))/dx#

We digress to use the chain rule:

Let #u = x+1#, then #(du)/dx = 1#

#(d(e^(x+1)))/dx = (d(e^u))/dx(du)/dx#

#(d(e^(x+1)))/dx = (e^u)(1)#

Reverse the substitution:

#(d(e^(x+1)))/dx = e^(x+1)#

Returning from the digression:

#dy/dx = e^(x+1)#