Question #93d8f

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Question #93d8f

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Answer 1 · 2018-01-06T05:30:07

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

Explanation:

.

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

Let $u = x + 1$ $u=x+1$

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

$y = e^{u}$ $y=e^u$

$\frac{d y}{d u} = e^{u}$ $dy/(du)=e^u$

Using the Chain Rule:

$\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}$ $dy/dx=dy/(du)*(du)/dx$

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

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

Answer 2 · 2018-01-06T11:32:20

$e^{x + 1}$ $e^(x+1)$

Explanation:

$differentiate using the chain rule$ $"differentiate using the "color(blue)"chain rule"$

$given y = f (g (x)) then$ $"given "y=f(g(x))" then "$

$dy/dx=f'(g(x))xxg'(x)$

$rArrd/dx(e^(x+1))=e^(x+1)xxd/dx(x+1)=e^(x+1)$

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Question #93d8f

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