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Question

Evaluate f(2)f^{\prime \prime \prime}(-2) where f(x)=2lnx+3f(x)=2 \ln |x|+3

Solution

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The function f(x)f(x) is given as

f(x)=2lnx+3\begin{align*} f(x) & = 2\, \ln |x| + 3 \end{align*}

Differentiating f(x)f(x) w.r.t. xx, we get

f(x)=21x+0f(x)=2x1\begin{align*} f'(x) & = 2\, \cdot \dfrac{1}{x} + 0 \\ f'(x) & = 2\, {x}^{-1} \end{align*}

Differentiating f(x)f'(x) w.r.t. xx, we get

f(x)=(1)2x2f(x)=2x2\begin{align*} f''(x) & = (-1)\,\cdot 2\, {x}^{-2} \\ f''(x) & = -2\, {x}^{-2} \\ \end{align*}

Differentiating f(x)f''(x) w.r.t. xx, we get

f(x)=(2)2x3f(x)=4x3\begin{align*} f'''(x) & = -(-2)\,\cdot 2\, {x}^{-3} \\ f'''(x) & = 4\, {x}^{-3} \\ \end{align*}

Therefore. the value of f(2)f'''(-2) is given by

f(2)=4(2)3f(2)=4(2)3f(2)=41(2)3f(2)=4(18)f(2)=12f(2)=12\begin{align*} f'''(-2) & = 4\, {(-2)}^{-3} \\ f'''(-2) & = 4\,\cdot {(-2)}^{-3} \\ f'''(-2) & = 4\, \cdot \dfrac{1}{(-2)^3}\\ f'''(-2) & = 4\, \cdot \left(\dfrac{1}{-8}\right)\\ f'''(-2) & = -\dfrac{1}{2}\\ &\hspace*{-14mm}\boxed{f'''(-2)= -\dfrac{1}{2}} \end{align*}

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