Question - Finding Second Derivative of Natural Logarithm Function
Solution:
The image shows a mathematical expression and a request to find the second derivative of the function with respect to x. The function is $$ y = \ln(2x^2 + x) $$. We need to find $$ \frac{d^2y}{dx^2} $$ when $$ x = -1 $$.First, let's find the first derivative $$ \frac{dy}{dx} $$. To do this, we'll use the chain rule since y is the natural logarithm of a function of x.Let $$ u = 2x^2 + x $$, so $$ y = \ln(u) $$.First, let's differentiate $$ u $$ with respect to $$ x $$:\[ \frac{du}{dx} = \frac{d(2x^2 + x)}{dx} = 4x + 1 \]Now let's differentiate $$ y $$ with respect to $$ u $$:\[ \frac{dy}{du} = \frac{d(\ln(u))}{du} = \frac{1}{u} \]Now use the chain rule to find $$ \frac{dy}{dx} $$:\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{u} \cdot (4x + 1) = \frac{4x + 1}{2x^2 + x} \]Next, we'll find the second derivative $$ \frac{d^2y}{dx^2} $$. For this, we need to differentiate $$ \frac{dy}{dx} $$ with respect to $$ x $$ using the quotient rule:Let $$ v = 4x + 1 $$ and $$ w = 2x^2 + x $$, so $$ v' = 4 $$ and $$ w' = 4x + 1 $$.Now apply the quotient rule $$ \left( \frac{v}{w} \right)' = \frac{v'w - vw'}{w^2} $$:\[ \frac{d^2y}{dx^2} = \frac{v'w - vw'}{w^2} = \frac{4(2x^2 + x) - (4x + 1)(4x + 1)}{(2x^2 + x)^2} \]Now simplify and find $$ \frac{d^2y}{dx^2} $$ when $$ x = -1 $$:\[ \frac{d^2y}{dx^2} = \frac{8x^2 + 4x - (16x^2 + 8x + 1)}{(2x^2 + x)^2} \]\[ \frac{d^2y}{dx^2} = \frac{8x^2 + 4x - 16x^2 - 8x - 1}{(2x^2 + x)^2} \]\[ \frac{d^2y}{dx^2} = \frac{-8x^2 - 4x - 1}{(2x^2 + x)^2} \]Now plug in $$ x = -1 $$:\[ \frac{d^2y}{dx^2}\Bigg|_{x=-1} = \frac{-8(-1)^2 - 4(-1) - 1}{(2(-1)^2 - 1)^2} \]\[ \frac{d^2y}{dx^2}\Bigg|_{x=-1} = \frac{-8 + 4 - 1}{(2 + (-1))^2} \]\[ \frac{d^2y}{dx^2}\Bigg|_{x=-1} = \frac{-5}{1^2} \]\[ \frac{d^2y}{dx^2}\Bigg|_{x=-1} = -5 \]The second derivative of the function at $$ x = -1 $$ is $$ -5 $$.
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