Question Statement

Let $f(x)=x^3+2x$.

a. Find $f^{\prime }(x)$ and $f^{\prime \prime }(x)$.

b. In general, $f^{\prime \prime }(x)={\lim }_{\Delta x\to 0}\frac{f^{\prime }(x+\Delta x)-f^{\prime }(x)}{\Delta x}$ provided the limit exists. Use $f^{\prime }(x)$ obtained in part (a) and use the definition to find $f^{\prime \prime }(x)$.

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Background and Explanation

This problem covers both standard differentiation techniques and the fundamental limit definition of derivatives. Part (a) applies the power rule for quick computation, while part (b) verifies the result using the formal definition of the derivative, reinforcing the conceptual foundation of calculus.

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Solution

Part (a)

Given the function: $f(x)=x^3+2x$

Finding the first derivative $f^{\prime }(x)$:

Differentiate term by term with respect to $x$ using the power rule $\frac{d}{dx}(x^n)=n{x}^{n-1}$:

$f^{\prime }(x)=\frac{d}{dx}(x^3)+2\frac{d}{dx}(x)$

$f^{\prime }(x)=3x^2+2$

Finding the second derivative $f^{\prime \prime }(x)$:

Differentiate $f^{\prime }(x)$ with respect to $x$:

$f^{\prime \prime }(x)=\frac{d}{dx}(3x^2+2)$

$f^{\prime \prime }(x)=3(2x)+0$

$f^{\prime \prime }(x)=6x$

Part (b)

Using the limit definition of the second derivative: $f^{\prime \prime }(x)={\lim }_{\Delta x\to 0}\frac{f^{\prime }(x+\Delta x)-f^{\prime }(x)}{\Delta x}$

From part (a), we know $f^{\prime }(x)=3x^2+2$.

Step 1: Find $f^{\prime }(x+\Delta x)$ by substituting $(x+\Delta x)$ into the first derivative: $f^{\prime }(x+\Delta x)=3(x+\Delta x{)}^2+2$

Expand $(x+\Delta x{)}^2$: $f^{\prime }(x+\Delta x)=3(x^2+2x\Delta x+(\Delta x{)}^2)+2$ $f^{\prime }(x+\Delta x)=3x^2+6x\Delta x+3(\Delta x{)}^2+2$

Step 2: Substitute into the limit definition: $f^{\prime \prime }(x)={\lim }_{\Delta x\to 0}\frac{[3(x+\Delta x{)}^2+2]-[3x^2+2]}{\Delta x}$

$={\lim }_{\Delta x\to 0}\frac{[3x^2+6x\Delta x+3(\Delta x{)}^2+2]-3x^2-2}{\Delta x}$

Step 3: Simplify the numerator: $={\lim }_{\Delta x\to 0}\frac{3x^2+6x\Delta x+3(\Delta x{)}^2+2-3x^2-2}{\Delta x}$

$={\lim }_{\Delta x\to 0}\frac{6x\Delta x+3(\Delta x{)}^2}{\Delta x}$

Step 4: Factor out $\Delta x$ from the numerator: $={\lim }_{\Delta x\to 0}\frac{\Delta x(6x+3\Delta x)}{\Delta x}$

Cancel $\Delta x$ from numerator and denominator: $={\lim }_{\Delta x\to 0}(6x+3\Delta x)$

Step 5: Evaluate the limit as $\Delta x\to 0$: $=6x+3(0)$

$f^{\prime \prime }(x)=6x$

This confirms the result obtained in part (a).

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Key Formulas or Methods Used

• Power Rule: $\frac{d}{dx}(x^n)=n{x}^{n-1}$
• Sum Rule: $\frac{d}{dx}[f(x)+g(x)]=f^{\prime }(x)+g^{\prime }(x)$
• Constant Multiple Rule: $\frac{d}{dx}[c\cdot f(x)]=c\cdot f^{\prime }(x)$
• Limit Definition of the Second Derivative: $f^{\prime \prime }(x)={\lim }_{\Delta x\to 0}\frac{f^{\prime }(x+\Delta x)-f^{\prime }(x)}{\Delta x}$
• Algebraic Expansion: $(x+\Delta x{)}^2=x^2+2x\Delta x+(\Delta x{)}^2$

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Summary of Steps

1. Part (a): Apply the power rule to $f(x)=x^3+2x$ to find $f^{\prime }(x)=3x^2+2$
2. Part (a): Differentiate $f^{\prime }(x)$ to get $f^{\prime \prime }(x)=6x$
3. Part (b): Substitute $f^{\prime }(x)=3x^2+2$ into the limit definition $f^{\prime \prime }(x)={\lim }_{\Delta x\to 0}\frac{f^{\prime }(x+\Delta x)-f^{\prime }(x)}{\Delta x}$
4. Part (b): Expand $f^{\prime }(x+\Delta x)=3(x+\Delta x{)}^2+2$ algebraically to $3x^2+6x\Delta x+3(\Delta x{)}^2+2$
5. Part (b): Subtract $f^{\prime }(x)$ and simplify the difference quotient to $6x+3\Delta x$
6. Part (b): Take the limit as $\Delta x\to 0$ to verify $f^{\prime \prime }(x)=6x$