Q. \[ \frac{d}{dx}\left(e^{x^2}\right) \]

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

Use chain rule: \dfrac{d}{dx} e^{x^2} = e^{x^2}\cdot \dfrac{d}{dx}(x^2)=2x e^{x^2}.

Q: How do you differentiate e^{x^2} using the chain rule step by step?

Let y=e^{u} with u=x^2. Then y'=e^{u}u'. Since u'=2x, y'=2x e^{x^2}.

Q: What is \dfrac{d}{dx}\left( e^{(x^2)} \right)?

The parentheses don’t change anything: exponent is still x^2. So \dfrac{d}{dx} e^{x^2}=2x e^{x^2}.

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

Chain rule again: \dfrac{d}{dx} e^{2x^2}=e^{2x^2}\cdot \dfrac{d}{dx}(2x^2)=e^{2x^2}\cdot 4x=4x e^{2x^2}.

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

First derivative: y'=2x e^{x^2}. Differentiate: y''=2e^{x^2}+2x\cdot e^{x^2}\cdot 2x=2e^{x^2}+4x^2 e^{x^2}=(4x^2+2)e^{x^2}.

Q: What is \dfrac{d}{dx}\left( \ln(e^{x^2}) \right)?

\ln(e^{x^2})=x^2. Then \dfrac{d}{dx}(x^2)=2x.

Q: How to differentiate e^{x^2+1}?

Since \dfrac{d}{dx}(x^2+1)=2x, \dfrac{d}{dx} e^{x^2+1}=e^{x^2+1}\cdot 2x=2x e^{x^2+1}.

Answer

We need the derivative of \(e^{x^2}\). Use the chain rule: if \(y=e^{u}\), then \(\frac{dy}{dx}=e^{u}\frac{du}{dx}\). Here \(u=x^2\), so \(\frac{du}{dx}=2x\).

\[ \frac{d}{dx}\left(e^{x^2}\right)=e^{x^2}\cdot 2x=2xe^{x^2}. \]

Detailed Explanation

We want to differentiate the function

\(f(x)=e^{x^2}\).

Step 1: Identify the outer and inner functions.

Use the chain rule. Think of \(f(x)\) as a composition:

Outer function: \(g(u)=e^{u}\)
Inner function: \(u=x^2\)

Step 2: Differentiate the outer function.

Differentiate \(g(u)=e^{u}\) with respect to \(u\):

\[
\frac{d}{du}e^{u}=e^{u}
\]

Step 3: Differentiate the inner function.

Now differentiate \(u=x^2\) with respect to \(x\):

\[
\frac{d}{dx}x^2=2x
\]

Step 4: Apply the chain rule.

The chain rule says:

\[
\frac{d}{dx}e^{x^2}=e^{x^2}\cdot 2x
\]

Final answer:

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

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Calculus FAQ

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

Use chain rule: \(\dfrac{d}{dx} e^{x^2} = e^{x^2}\cdot \dfrac{d}{dx}(x^2)=2x e^{x^2}\).

How do you differentiate \(e^{x^2}\) using the chain rule step by step?

Let \(y=e^{u}\) with \(u=x^2\). Then \(y'=e^{u}u'\). Since \(u'=2x\), \(y'=2x e^{x^2}\).

What is \(\dfrac{d}{dx}\left( e^{(x^2)} \right)\)?

The parentheses don’t change anything: exponent is still \(x^2\). So \(\dfrac{d}{dx} e^{x^2}=2x e^{x^2}\).

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

Chain rule again: \(\dfrac{d}{dx} e^{2x^2}=e^{2x^2}\cdot \dfrac{d}{dx}(2x^2)=e^{2x^2}\cdot 4x=4x e^{2x^2}\).

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

First derivative: \(y'=2x e^{x^2}\). Differentiate: \(y''=2e^{x^2}+2x\cdot e^{x^2}\cdot 2x=2e^{x^2}+4x^2 e^{x^2}=(4x^2+2)e^{x^2}\).

What is \(\dfrac{d}{dx}\left( \ln(e^{x^2}) \right)\)?

\(\ln(e^{x^2})=x^2\). Then \(\dfrac{d}{dx}(x^2)=2x\).

How to differentiate \(e^{x^2+1}\)?

Since \(\dfrac{d}{dx}(x^2+1)=2x\), \(\dfrac{d}{dx} e^{x^2+1}=e^{x^2+1}\cdot 2x=2x e^{x^2+1}\).

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