Question Statement

Find the first and second derivatives of the function:

$f(x)=e^{2x}\left(x^2+1\right)$

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Background and Explanation

This problem requires applying the product rule for differentiation combined with the chain rule to handle the composite exponential function $e^{2x}$. You will differentiate successively to obtain both the first and second derivatives.

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Solution

Finding the First Derivative $f^{\prime }(x)$

We apply the product rule $\frac{d}{dx}[u\cdot v]=u^{\prime }\cdot v+u\cdot v^{\prime }$ where $u=e^{2x}$ and $v=x^2+1$.

First, compute the derivatives of the individual components using the chain rule for the exponential:

• $\frac{d}{dx}(e^{2x})=e^{2x}\cdot 2=2e^{2x}$
• $\frac{d}{dx}(x^2+1)=2x$

Now apply the product rule:

$\begin{array}{rl}{f}^{\prime }(x)& =\frac{d}{dx}\left(e^{2x}\right)\cdot \left(x^2+1\right)+e^{2x}\cdot \frac{d}{dx}\left(x^2+1\right)\\ & =e^{2x}\cdot 2\cdot \left(x^2+1\right)+e^{2x}\cdot (2x+0)\\ & =2e^{2x}(x^2+1)+2x{e}^{2x}\end{array}$

Factor out the common term $2e^{2x}$:

$f^{\prime }(x)=2e^{2x}\left[x^2+1+x\right]=2e^{2x}(x^2+x+1)$

Finding the Second Derivative $f^{\prime \prime }(x)$

Now differentiate $f^{\prime }(x)=2e^{2x}(x^2+x+1)$ with respect to $x$. We again use the product rule, treating the constant $2$ as a coefficient:

$\begin{array}{rl}{f}^{\prime \prime }(x)& =2\left[\frac{d}{dx}\left(e^{2x}\right)\cdot \left(x^2+x+1\right)+e^{2x}\cdot \frac{d}{dx}\left(x^2+x+1\right)\right]\\ & =2\left[e^{2x}\cdot 2\left(x^2+x+1\right)+e^{2x}\cdot (2x+1+0)\right]\\ & =2e^{2x}\left[2(x^2+x+1)+(2x+1)\right]\\ & =2e^{2x}\left[2x^2+2x+2+2x+1\right]\\ f^{\prime \prime }(x)& =2e^{2x}\left(2x^2+4x+3\right)\end{array}$

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Key Formulas or Methods Used

• Product Rule: $\frac{d}{dx}[u(x)v(x)]=u^{\prime }(x)v(x)+u(x)v^{\prime }(x)$
• Chain Rule: $\frac{d}{dx}[e^{u(x)}]=e^{u(x)}\cdot u^{\prime }(x)$
• Power Rule: $\frac{d}{dx}[x^n]=n{x}^{n-1}$

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Summary of Steps

1. Identify components: Recognize $f(x)$ as a product of $e^{2x}$ and $(x^2+1)$
2. First differentiation: Apply product rule to find $f^{\prime }(x)=2e^{2x}(x^2+1)+2x{e}^{2x}$
3. Simplify $f^{\prime }(x)$: Factor to get $f^{\prime }(x)=2e^{2x}(x^2+x+1)$
4. Second differentiation: Apply product rule again to $f^{\prime }(x)$, differentiating $e^{2x}(x^2+x+1)$
5. Simplify $f^{\prime \prime }(x)$: Combine like terms inside the brackets to obtain $f^{\prime \prime }(x)=2e^{2x}(2x^2+4x+3)$