Q. \(\displaystyle \frac{d}{dx}\left(\frac{1}{e^x}\right)\)

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

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

Q: Can I use the quotient rule for e^{-x} ?

Yes, but it’s unnecessary. Using e^{-x} is simplest: chain rule gives -e^{-x} . Since \frac{1}{e^x} is a single exponential, quotient rule is more work.

Q: What is the derivative of e^{-x} directly?

By the chain rule, \frac{d}{dx}(e^{g(x)})=e^{g(x)}g'(x) . Here g(x)=-x , so \frac{d}{dx}(e^{-x})=e^{-x}(-1)=-e^{-x} .

Q: What is \frac{d}{dx}\left(e^x\right) and how does it relate?

\frac{d}{dx}(e^x)=e^x . Since \frac{1}{e^x}=e^{-x} , the exponent flips sign, producing -e^{-x} .

Q: How do I differentiate \frac{1}{e^x} using limits?

Let f(x)=e^{-x} . Then f'(x)=\lim_{h\to 0}\frac{e^{-(x+h)}-e^{-x}}{h}=e^{-x}\lim_{h\to 0}\frac{e^{-h}-1}{h}=-e^{-x} .

Q: What is the derivative of \frac{1}{e^{2x}} ?

\frac{1}{e^{2x}}=e^{-2x} . Chain rule gives \frac{d}{dx}(e^{-2x})=e^{-2x}(-2)=-2e^{-2x}=-\frac{2}{e^{2x}} .

Answer

We use \( \frac{d}{dx}\left(e^x\right)=e^x \) and the derivative of an inverse/power.

Rewrite:

\[
\frac{1}{e^x} = e^{-x}
\]

Differentiate:

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

Final result:

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

Detailed Explanation

We want to find the derivative of the function

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

Step 1: Rewrite the function using exponent rules

Recall that

\[
\frac{1}{e^x}=e^{-x}.
\]

So our function can be written as

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

Step 2: Differentiate using the chain rule

Recall the derivative rule:

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

Here, \(u(x)=-x\). First find \(u'(x)\):

\[
u(x)=-x \quad \Rightarrow \quad u'(x)=-1.
\]

Now differentiate \(f(x)=e^{u(x)}\):

\[
f'(x)=e^{u(x)}\cdot u'(x)=e^{-x}\cdot(-1)=-e^{-x}.
\]

Step 3: Rewrite in the original form (optional)

Since \(e^{-x}=\frac{1}{e^x}\), we can write

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

Final Answer

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

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Calculus FAQ

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

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

Can I use the quotient rule for \( e^{-x} \)?

Yes, but it’s unnecessary. Using \( e^{-x} \) is simplest: chain rule gives \( -e^{-x} \). Since \( \frac{1}{e^x} \) is a single exponential, quotient rule is more work.

What is the derivative of \( e^{-x} \) directly?

By the chain rule, \( \frac{d}{dx}(e^{g(x)})=e^{g(x)}g'(x) \). Here \( g(x)=-x \), so \( \frac{d}{dx}(e^{-x})=e^{-x}(-1)=-e^{-x} \).

What is \( \frac{d}{dx}\left(e^x\right) \) and how does it relate?

\( \frac{d}{dx}(e^x)=e^x \). Since \( \frac{1}{e^x}=e^{-x} \), the exponent flips sign, producing \( -e^{-x} \).

How do I differentiate \( \frac{1}{e^x} \) using limits?

Let \( f(x)=e^{-x} \). Then \( f'(x)=\lim_{h\to 0}\frac{e^{-(x+h)}-e^{-x}}{h}=e^{-x}\lim_{h\to 0}\frac{e^{-h}-1}{h}=-e^{-x} \).

What is the derivative of \( \frac{1}{e^{2x}} \)?

\( \frac{1}{e^{2x}}=e^{-2x} \). Chain rule gives \( \frac{d}{dx}(e^{-2x})=e^{-2x}(-2)=-2e^{-2x}=-\frac{2}{e^{2x}} \).

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