Single Variable Calculus, Chapter 3, 3.3, Section 3.3, Problem 59

Find the 1st and 2nd derivatives of $\displaystyle f(x) = \frac{x^2}{ 1 + 2x}$


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\begin{equation}
\begin{aligned}

f'(x) =& \frac{\displaystyle (1 + 2x) \frac{d}{dx} (x^2) - \left[ (x^2) \frac{d}{dx} (1 + 2x) \right] }{(1 + 2x)^2}
&& \text{Using Quotient Rule}
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f'(x) =& \frac{(1 + 2x)(2x) - (x^2)(2)}{(1 + 2x)^2}
&& \text{Simplify the equation}
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f'(x) =& \frac{2x + 4x^2 - 2x^2}{(1 + 2x)^2}
&& \text{Combine like terms}
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f'(x) =& \frac{2x^2 + 2x}{(1 + 2x)^2}
&& \text{1st derivative of $f(x)$}
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f''(x) =& \frac{\displaystyle (1 + 4x + 4x^2) \frac{d}{dx} (2x^2 + 2x) - \left[ (2x^2 + 2x) \frac{d}{dx} (1 + 4x + 4x^2) \right]}{[(1 + 2x)^2]^2}
&& \text{Using Quotient Rule}
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f''(x) =& \frac{(1 + 4x + 4x^2) (4x + 2) - [(2x^2 + 2x)(0 + 4 + 8x)]}{(1 + 2x)^4}
&& \text{Simplify the equation}
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f''(x) =& \frac{4x + \cancel{16x^2} + \cancel{16x^3} + 2 + \cancel{8x} + \cancel{8x^2} - \cancel{8x^2} - \cancel{16x^3} - \cancel{8x} - \cancel{16x^2}}{(1 + 2x)^4}
&& \text{Combine like terms}
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f''(x) =& \frac{4x + 2}{(1 + 2x )^4}
&& \text{Simplify the equation}
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f''(x) =& \frac{2 \cancel{(2x + 1)}}{\cancel{(2x + 1)}(2x + 1)^3}
&& \text{Factor the equation}
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f''(x) =& \frac{2}{(2x + 1)^3}
&& \text{2nd derivative of $f(x)$}
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\end{aligned}
\end{equation}
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