How do you differentiate $y = e^{x + 1} + 1$ $y=e^(x+1)+1$ ?

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Answer 1 · 2017-07-10T17:15:08

Given: $y = e^{x + 1} + 1$ $y=e^(x+1)+1$

Differentiate each term:

$\frac{d y}{d x} = \frac{d (e^{x + 1})}{d x} + \frac{d (1)}{d x}$ $dy/dx = (d(e^(x+1)))/dx + (d(1))/dx$

The derivative of a constant is 0:

$\frac{d y}{d x} = \frac{d (e^{x + 1})}{d x} + 0$ $dy/dx = (d(e^(x+1)))/dx + 0$

$\frac{d y}{d x} = \frac{d (e^{x + 1})}{d x}$ $dy/dx = (d(e^(x+1)))/dx$

We digress to use the chain rule:

Let $u = x + 1$ $u = x+1$ , then $\frac{d u}{d x} = 1$ $(du)/dx = 1$

$\frac{d (e^{x + 1})}{d x} = \frac{d (e^{u})}{d x} \frac{d u}{d x}$ $(d(e^(x+1)))/dx = (d(e^u))/dx(du)/dx$

$\frac{d (e^{x + 1})}{d x} = (e^{u}) (1)$ $(d(e^(x+1)))/dx = (e^u)(1)$

Reverse the substitution:

$\frac{d (e^{x + 1})}{d x} = e^{x + 1}$ $(d(e^(x+1)))/dx = e^(x+1)$

Returning from the digression:

$\frac{d y}{d x} = e^{x + 1}$ $dy/dx = e^(x+1)$

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