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Chapter 3: Problem 71

Find the following higher-order derivatives. $$\frac{d^{2}}{d x^{2}}\left(\log _{10} x\right)$$

Short Answer

Expert verified
Answer: The second-order derivative is \(\frac{-1}{\ln(10)x^2}\).

Step by step solution

01

Find the first derivative

To find the first derivative, we will use the change of base formula for logarithms. For a given function, \(y = \log_bx\), the change of base formula is $$y = \frac{\log_cx}{\log_cb}$$ where \(c\) is a new base. In our case, let's use the natural logarithm (\(c=e\)) for simplicity. So, we have: $$y = \log_{10}(x) = \frac{\ln(x)}{\ln(10)}$$ Now, we differentiate \(y\) with respect to \(x\) using the chain rule: $$\frac{dy}{dx} = \frac{d}{dx}\left(\frac{\ln(x)}{\ln(10)}\right) = \frac{1}{\ln(10)}\frac{d}{dx}\left(\ln(x)\right) = \frac{1}{\ln(10)}\cdot\frac{1}{x}$$
02

Find the second derivative

Now that we have the first derivative, we can differentiate it once more to find the second derivative: $$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{1}{\ln(10)}\cdot\frac{1}{x}\right)$$ Factor out the constant \(\frac{1}{\ln(10)}\): $$\frac{d^2y}{dx^2} = \frac{1}{\ln(10)}\cdot\frac{d}{dx}\left(\frac{1}{x}\right)$$ Using the power rule, rewrite \(\frac{1}{x}\) as \(x^{-1}\): $$\frac{d^2y}{dx^2} = \frac{1}{\ln(10)}\cdot\frac{d}{dx}\left(x^{-1}\right)$$ Now compute the derivative of \(x^{-1}\): $$\frac{d^2y}{dx^2} = \frac{1}{\ln(10)}\cdot\left(-1\cdot x^{-2}\right)$$ Finally, rewrite the derivative back in its original form: $$\frac{d^2y}{dx^2} = \frac{-1}{\ln(10)x^2}$$

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