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Problem 10

# Calculate $$\frac{d^{2} y}{d x^{2}}$$. . $$y=x^{-2}+\ln x$$

Expert verified
The second derivative of the given function, expressed as $$\frac{d^2 y}{d x^2}$$, is: $\frac{d^2 y}{d x^2} = \frac{6}{x^4} - \frac{1}{x^2}.$
See the step by step solution

## Step 1: Differentiate the given function y with respect to x

To find the first derivative of $$y = x^{-2} + \ln x$$, we will differentiate each term with respect to x. For the first term, $$x^{-2}$$, we apply the power rule for differentiation: $$\frac{d}{dx}(x^n) = nx^{n-1}$$. Thus, $\frac{d}{dx}(x^{-2}) = -2x^{-3}.$ For the second term, $$\ln x$$, the derivative of the natural logarithmic function is: $\frac{d}{dx}(\ln x) = \frac{1}{x}.$ Now, adding both derivative terms, we find the first derivative, $$y'$$: $y' = -2x^{-3} + \frac{1}{x}.$

## Step 2: Differentiate the 1st derivative for 2nd derivative

Having found the first derivative of the function, we will now differentiate it to find the second derivative $$\frac{d^2 y}{d x^2}$$. Differentiating each term separately, we get: For the first term, $$-2x^{-3}$$, we apply the power rule again: $\frac{d}{dx}(-2x^{-3}) = 6x^{-4}.$ For the second term, $$\frac{1}{x}$$, the derivative is: $\frac{d}{dx}\left(\frac{1}{x}\right) = -\frac{1}{x^2}.$ Now, adding both derivative terms, we find the second derivative, $$y''$$: $y'' = 6x^{-4} - \frac{1}{x^2}.$

## Step 3: Write the final answer in a simplified form

Finally, to express the second derivative in a more simplified form, we can rewrite the $$x^{-4}$$ term as $$\frac{1}{x^4}$$: $y'' = 6\frac{1}{x^4} - \frac{1}{x^2}.$ So, the second derivative of the given function, expressed as $$\frac{d^2 y}{d x^2}$$, is: $\frac{d^2 y}{d x^2} = \frac{6}{x^4} - \frac{1}{x^2}.$

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