Find the derivative of p(x). p(x) = ln(sqrt(x)) p′(x) = ______

Understand Function: Understand the function and what is being asked. We need to find the derivative of the function p(x) = ln(sqrt(x)). The derivative of a function gives us the rate at which the function's value changes with respect to changes in the variable x.Apply Chain Rule: Apply the chain rule for differentiation. The chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function. Here, the outer function is ln(u) and the inner function is sqrt(x) or x^(1/2).Differentiate Outer Function: Differentiate the outer function. The derivative of ln(u) with respect to u is 1/u. So, we have 1/sqrt(x) for the outer derivative when we substitute back the inner function.Differentiate Inner Function: Differentiate the inner function. The derivative of sqrt(x) or x^(1/2) with respect to x is (1/2)x^(-1/2).Multiply Derivatives: Multiply the derivatives from Step 3 and Step 4. We multiply 1/sqrt(x) by (1/2)x^(-1/2) to get the derivative of the composite function. This simplifies to (1/2) * (1/x^(1/2)) * x^(-1/2).Simplify Expression: Simplify the expression. We can simplify (1/2) * (1/x^(1/2)) * x^(-1/2) by combining the exponents of x. This results in (1/2) * x^(-1) or (1/2) * (1/x).Write Final Answer: Write the final answer. The derivative of p(x) = ln(sqrt(x)) is p′(x) = (1/2) * (1/x).

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Find the derivative of $p(x)$ . $\newline$ $p(x) = \ln(\sqrt{x})$ $\newline$ $p' (x) =$ ______

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Algebra 1

Domain and range of square root functions: equations

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Q. Find the derivative of

p(x)

\newline

p(x) = \ln(\sqrt{x})

\newline

p' (x) =

______

Understand Function: Understand the function and what is being asked. $\newline$ We need to find the derivative of the function $p(x) = \ln(\sqrt{x})$ . The derivative of a function gives us the rate at which the function's value changes with respect to changes in the variable $x$ .
Apply Chain Rule: Apply the chain rule for differentiation. The chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function. Here, the outer function is $\ln(u)$ and the inner function is $\sqrt{x}$ or $x^{(1/2)}$ .
Differentiate Outer Function: Differentiate the outer function. $\newline$ The derivative of $\ln(u)$ with respect to $u$ is $\frac{1}{u}$ . So, we have $\frac{1}{\sqrt{x}}$ for the outer derivative when we substitute back the inner function.
Differentiate Inner Function: Differentiate the inner function. The derivative of $\sqrt{x}$ or $x^{(1/2)}$ with respect to $x$ is $(1/2)x^{(-1/2)}$ .
Multiply Derivatives: Multiply the derivatives from Step $3$ and Step $4$ . $\newline$ We multiply $\frac{1}{\sqrt{x}}$ by $(\frac{1}{2})x^{(-\frac{1}{2})}$ to get the derivative of the composite function. This simplifies to $(\frac{1}{2}) \cdot (\frac{1}{x^{(\frac{1}{2})}}) \cdot x^{(-\frac{1}{2})}.$
Simplify Expression: Simplify the expression. $\newline$ We can simplify $(\frac{1}{2}) \times (\frac{1}{x^{\frac{1}{2}}}) \times x^{-\frac{1}{2}}$ by combining the exponents of $x$ . This results in $(\frac{1}{2}) \times x^{-1}$ or $(\frac{1}{2}) \times (\frac{1}{x})$ .
Write Final Answer: Write the final answer. $\newline$ The derivative of $p(x) = \ln(\sqrt{x})$ is $p′(x) = \frac{1}{2} \cdot \frac{1}{x}$ .