Question

In Exercises \(37-42,\) find \(f^{\prime}(x)\) and state the domain of \(f^{\prime}\) $$f(x)=\log _{10} \sqrt{x+1}$$

Short answer

The derivative \(f^{\prime}(x)\) is \(\frac{1}{2} \cdot \frac{1}{(x+1) \ln 10}\) and the domain of \(f^{\prime}(x)\) is \(-\infty < x < -1\) or \(-1 < x < \infty\).

Step-by-step solution

Rewrite the function using the property of logarithms

We begin by changing f(x) slightly to make it easier to deal with. We use the property of logarithms \(\log_b(a^n)=n\log_b(a)\) to simplify the function: \(f(x)=\frac{1}{2} \log_{10}(x+1)\). This will make the differentiation step easier.

Differentiate using the chain rule

Next, we differentiate the function. The derivative of \(\log_{10}(a)\) is \(\frac{1}{a \ln 10}\), and because of the chain rule, we will multiply it by the derivative of \(x+1\) which is \(1\). So the derivative \(f'(x)\) is: \[f^{\prime}(x)=\frac{1}{2} \cdot \frac{1}{(x+1) \ln 10}\]

Find the domain of the derivative function

To determine the domain of the derivative function, we need to know when the derivative is defined. The derivative is not defined when the denominator equals 0, that is, when \(x+1=0\), or \(x=-1\). Hence, the derivative function is defined for all real numbers except \(-1\). Therefore, the domain of \(f^{\prime}(x)\) is \(-\infty < x < -1\) or \(-1 < x < \infty\).