Q. Find the derivative of \(e^{\frac{1}{x}}\).

Q: What is the derivative of e^{1/x}?

f(x)=e^{1/x}. Then f'(x)=e^{1/x}\cdot \frac{d}{dx}(1/x)=e^{1/x}\cdot\left(-\frac{1}{x^2}\right)=-\frac{e^{1/x}}{x^2}.

Q: Why do we use the chain rule here?

Because e^{1/x} is a composite function: outer e^u with u=1/x. Chain rule: \frac{d}{dx}e^{u}=e^{u}\cdot u'.

Q: What is \frac{d}{dx}\left(\frac{1}{x}\right)?

\frac{d}{dx}\left(x^{-1}\right)=-x^{-2}=-\frac{1}{x^2}.

Q: What is the derivative of e^{a/x} for constant a?

Let u=a/x. Then u'=-a/x^2. So \frac{d}{dx}e^{a/x}=e^{a/x}\cdot\left(-\frac{a}{x^2}\right)=-\frac{a e^{a/x}}{x^2}.

Q: What is the derivative of \left(e^{1/x}\right)^{-1} or e^{-1/x}?

For e^{-1/x}, use u=-1/x, so u'=1/x^2. Thus \frac{d}{dx}e^{-1/x}=e^{-1/x}\cdot\frac{1}{x^2}=\frac{e^{-1/x}}{x^2}.

Q: What is the domain of the derivative?

e^{1/x} is undefined at x=0. The derivative -\frac{e^{1/x}}{x^2} is valid for all x\neq 0.

Q: What is the second derivative of e^{1/x}?

From f'(x)=-e^{1/x}x^{-2}. Differentiate: f''(x)=e^{1/x}\left(\frac{2}{x^3}+\frac{1}{x^4}\right)=\frac{e^{1/x}}{x^4}(2x+1).

Answer

Let \(f(x)=e^{1/x}\). Using the chain rule, the derivative is the derivative of the outer function times the derivative of the inner function:

\[
\frac{d}{dx}\left(e^{1/x}\right)=e^{1/x}\cdot \frac{d}{dx}\left(\frac{1}{x}\right)
\]
\[
\frac{d}{dx}\left(\frac{1}{x}\right)=-\frac{1}{x^2}
\]

So, the final result is:

\[
\frac{d}{dx}\left(e^{1/x}\right)=-\frac{e^{1/x}}{x^2}
\]

Detailed Explanation

We want to find the derivative of the function

\[
f(x)=e^{\frac{1}{x}}.
\]

This is a composite function, so we will use the Chain Rule.

Step 1: Identify the outer and inner functions.

Rewrite the expression as

\[
f(x)=\exp\!\left(g(x)\right)
\]

where the outer function is

\[
\exp(u)=e^{u}
\]

and the inner function is

\[
g(x)=\frac{1}{x}.
\]

Step 2: Differentiate the outer function.

The derivative of \(e^{u}\) with respect to \(u\) is

\[
\frac{d}{du}\left(e^{u}\right)=e^{u}.
\]

Step 3: Differentiate the inner function.

Now compute the derivative of \(g(x)=\frac{1}{x}=x^{-1}\):

\[
g'(x)=\frac{d}{dx}\left(x^{-1}\right)=-x^{-2}=-\frac{1}{x^{2}}.
\]

Step 4: Apply the Chain Rule.

The Chain Rule says:

\[
\frac{d}{dx}\left(e^{g(x)}\right)=e^{g(x)}\cdot g'(x).
\]

Substitute \(g(x)=\frac{1}{x}\) and \(g'(x)=-\frac{1}{x^{2}}\):

\[
f'(x)=e^{\frac{1}{x}}\left(-\frac{1}{x^{2}}\right).
\]

Final Answer:

\[
\frac{d}{dx}\left(e^{\frac{1}{x}}\right)=-\frac{1}{x^{2}}\,e^{\frac{1}{x}}.
\]

See full solution

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Calculus FAQ

What is the derivative of \(e^{1/x}\)?

\(f(x)=e^{1/x}\). Then \(f'(x)=e^{1/x}\cdot \frac{d}{dx}(1/x)=e^{1/x}\cdot\left(-\frac{1}{x^2}\right)=-\frac{e^{1/x}}{x^2}\).

Why do we use the chain rule here?

Because \(e^{1/x}\) is a composite function: outer \(e^u\) with \(u=1/x\). Chain rule: \(\frac{d}{dx}e^{u}=e^{u}\cdot u'\).

What is \(\frac{d}{dx}\left(\frac{1}{x}\right)\)?

\(\frac{d}{dx}\left(x^{-1}\right)=-x^{-2}=-\frac{1}{x^2}\).

What is the derivative of \(e^{a/x}\) for constant \(a\)?

Let \(u=a/x\). Then \(u'=-a/x^2\). So \( \frac{d}{dx}e^{a/x}=e^{a/x}\cdot\left(-\frac{a}{x^2}\right)=-\frac{a e^{a/x}}{x^2}\).

What is the derivative of \(\left(e^{1/x}\right)^{-1}\) or \(e^{-1/x}\)?

For \(e^{-1/x}\), use \(u=-1/x\), so \(u'=1/x^2\). Thus \(\frac{d}{dx}e^{-1/x}=e^{-1/x}\cdot\frac{1}{x^2}=\frac{e^{-1/x}}{x^2}\).

What is the domain of the derivative?

\(e^{1/x}\) is undefined at \(x=0\). The derivative \(-\frac{e^{1/x}}{x^2}\) is valid for all \(x\neq 0\).

What is the second derivative of \(e^{1/x}\)?

From \(f'(x)=-e^{1/x}x^{-2}\). Differentiate: \(f''(x)=e^{1/x}\left(\frac{2}{x^3}+\frac{1}{x^4}\right)=\frac{e^{1/x}}{x^4}(2x+1)\).

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