How do you differentiate $y = e^{- \frac{x}{4}}$ $y=e^(-x/4)$ ?

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How do you differentiate $y = e^{- \frac{x}{4}}$ $y=e^(-x/4)$ ?

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Answer 1 · 2018-03-04T20:42:42

$\frac{d y}{d x} = - \frac{e^{- \frac{x}{4}}}{4}$ $dy/dx=-e^(-x/4)/4$

Explanation:

In order to differentiate a function of a function, we use the chain rule. In other words, if $y = f (u)$ $y=f(u)$ and $u = f (x)$ $u=f(x)$ , then

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

In this case, let $u = - \frac{x}{4}$ $u=-x/4$ and $y = e^{u}$ $y=e^u$ . Then

$\frac{d y}{d u} = e^{u} = e^{- \frac{x}{4}}$ $dy/(du)=e^u=e^(-x/4)$ ;

$\frac{d u}{d x} = - \frac{1}{4}$ $(du)/dx=-1/4$

and, importantly,

$\frac{d y}{d x} = \frac{d y}{d u} \frac{d u}{d x} = e^{- \frac{x}{4}} (- \frac{1}{4}) = - \frac{e^{- \frac{x}{4}}}{4}$ $dy/dx=dy/(du)(du)/dx=e^(-x/4)(-1/4)=-e^(-x/4)/4$

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How do you differentiate $y = e^{- \frac{x}{4}}$ $y=e^(-x/4)$ ?

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