Finding Second Derivative of Natural Logarithm Function
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 \( -5 \).
