How do you find the derivative of $2 e^{- x}$ $2e^-x$ ?

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Answer 1 · 2016-10-31T05:15:42

$\frac{d y}{d x} = - 2 e^{- x}$ $(dy)/(dx)=-2e^-x$

Recall that $\frac{d}{d x} e^{x} = e^{x}$ $d/dxe^x=e^x$

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

Let $u = - x$ $u=-x$

$\frac{d y}{d u} = \frac{d}{d u} 2 e^{u} = 2 e^{u} = 2 e^{- x}$ $(dy)/(du)=d/(du)2e^u=2e^u=2e^-x$

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

$:.(dy)/(dx)=-1*2e^-x=-2e^-x$

How do you find the derivative of 2e^-x2e−x2e^-x?