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

Q: What is \frac{d}{dx} e^{-x} ?

Use the chain rule: \frac{d}{dx} e^{-x} = e^{-x}\cdot \frac{d}{dx}(-x)= -e^{-x}.

Q: How do you differentiate e^{ax} for a constant a?

\frac{d}{dx} e^{ax} = a e^{ax}, since \frac{d}{dx}(ax)=a.

Q: What is \frac{d}{dx} e^{-kx} for constant k?

Chain rule gives \frac{d}{dx} e^{-kx} = e^{-kx}\cdot (-k)= -k e^{-kx}.

Q: What is the derivative of e^{-x}+e^{-2x} ?

Differentiate termwise: \frac{d}{dx}(e^{-x})=-e^{-x} and \frac{d}{dx}(e^{-2x})=-2e^{-2x}. Sum: -e^{-x}-2e^{-2x}.

Q: What is \frac{d}{dx} \big(3e^{-x}\big) ?

Constant multiple rule: \frac{d}{dx}\big(3e^{-x}\big)=3\cdot(-e^{-x})=-3e^{-x}.

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

First derivative: -e^{-x}. Differentiate again: \frac{d}{dx}(-e^{-x})= -(-e^{-x})= e^{-x}.

Answer

To differentiate \(e^{-x}\), use the chain rule.

Let the inside function be \(u=-x\). Then \( \frac{d}{dx} e^{u} = e^{u}\cdot \frac{du}{dx}\).

\(\frac{du}{dx} = -1\), so

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

Final result: \(-e^{-x}\).

Detailed Explanation

We want to find the derivative of the function

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

Step 1: Identify the outside function and the inside function (Chain Rule).

The Chain Rule applies to compositions of the form \(f(g(x))\).

Here:

The outside function is \(e^{u}\).
The inside function is \(u=-x\).

So we can rewrite \(e^{-x}\) as \(e^{u}\) where \(u=-x\).

Step 2: Differentiate the outside function.

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

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

Step 3: Differentiate the inside function.

The inside function is \(u=-x\), so

\[
\frac{d}{dx}(-x)=-1.
\]

Step 4: Apply the Chain Rule.

The Chain Rule says:

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

Substitute \(u=-x\):

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

Step 5: Simplify.

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

Final Answer:

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

See full solution

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Calculus FAQ

What is \( \frac{d}{dx} e^{-x} \)?

Use the chain rule: \( \frac{d}{dx} e^{-x} = e^{-x}\cdot \frac{d}{dx}(-x)= -e^{-x}\).

How do you differentiate \( e^{ax} \) for a constant \(a\)?

\( \frac{d}{dx} e^{ax} = a e^{ax}\), since \( \frac{d}{dx}(ax)=a\).

What is \( \frac{d}{dx} e^{-kx} \) for constant \(k\)?

Chain rule gives \( \frac{d}{dx} e^{-kx} = e^{-kx}\cdot (-k)= -k e^{-kx}\).

What is the derivative of \( e^{-x}+e^{-2x} \)?

Differentiate termwise: \( \frac{d}{dx}(e^{-x})=-e^{-x}\) and \( \frac{d}{dx}(e^{-2x})=-2e^{-2x}\). Sum: \( -e^{-x}-2e^{-2x}\).

What is \( \frac{d}{dx} \big(3e^{-x}\big) \)?

Constant multiple rule: \( \frac{d}{dx}\big(3e^{-x}\big)=3\cdot(-e^{-x})=-3e^{-x}\).

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

First derivative: \( -e^{-x}\). Differentiate again: \( \frac{d}{dx}(-e^{-x})= -(-e^{-x})= e^{-x}\).

Solve the e^-x derivative now.
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