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Chapter 2: Problem 55

Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, give an example to show why it is false. $$ \frac{d}{d x} \log _{a} \sqrt{x}=\frac{1}{(\ln a) \sqrt{x}} $$

Short Answer

Expert verified
The statement is false. The correct derivative of \(f(x) = log_a\sqrt{x}\) is \(\frac{1}{2(\ln a)x}\), not \(\frac{1}{(\ln a) \sqrt{x}}\).

Step by step solution

01

Rewrite the function using logarithmic properties

To simplify the problem, let's rewrite the given function using logarithmic properties: Given function: \(f(x) = log_a\sqrt{x}\) Using the property \(log_{a}{x^n} = n\cdot log_{a}{x}\), we get: \(f(x) = \frac{1}{2} \cdot log_a{x}\)
02

Apply the Chain Rule

To find the derivative of \(f(x) = \frac{1}{2} \cdot log_a{x}\), we will use the chain rule. The chain rule states that: \( (\frac{d}{d x} f(g(x)) = f'(g(x)) \cdot g'(x) \). Here, \(f(u) = \frac{1}{2} \cdot log_a{u}\) and the inner function, \(g(x) = x\). We will differentiate both functions and then compute the derivative using the chain rule.
03

Differentiate the outer function

Let's differentiate the outer function, \(f(u) = \frac{1}{2} \cdot log_a{u}\): \(f'(u) = \frac{d}{d u} (\frac{1}{2} \cdot log_a{u}) = \frac{1}{2} \cdot \frac{d}{d u} (log_{a} u)\) Use the property \(\frac{d}{d u} (log_{a} u) = \frac{1}{u\ln a}\): \(f'(u) = \frac{1}{2} \cdot \frac{1}{u\ln a} = \frac{1}{2(\ln a)u}\)
04

Differentiate the inner function

Since the inner function is \(g(x) = x\), the derivative of the inner function is simply: \(g'(x) = \frac{d}{d x}(x) = 1\)
05

Apply the Chain Rule

Now, we apply the chain rule to find the derivative of the original function \(f(x) = \frac{1}{2} \cdot log_{a}{x}\): \(\frac{d}{d x} f(x) = f'(g(x)) \cdot g'(x) = (\frac{1}{2(\ln a)g(x)}) \cdot 1\) Substitute \(g(x) = x\) back into the equation: \(\frac{d}{d x} f(x) = \frac{1}{2(\ln a)x}\)
06

Compare with the given expression

We found the derivative as \(\frac{1}{2(\ln a)x}\). Let's compare it with the given expression \(\frac{1}{(\ln a) \sqrt{x}}\): Given expression: \(\frac{1}{(\ln a) \sqrt{x}} = \frac{1}{2 (\ln a)\sqrt{x}^2}\) Comparing both expressions, we see that they are not equal. Therefore, the statement is false.
07

Provide a counter-example

Since the statement is false, let's provide a counter-example: Original function: \(f(x) = log_a\sqrt{x}\) Derivative (calculated): \(\frac{d}{d x} f(x) = \frac{1}{2(\ln a)x}\) Given derivative (false): \(\frac{1}{(\ln a) \sqrt{x}}\) In conclusion, the given statement is false, and the correct derivative of \(f(x) = log_a\sqrt{x}\) is \(\frac{1}{2(\ln a)x}\).

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Most popular questions from this chapter

Find the matrix \(A\) if $$ A\left[\begin{array}{rr} 1 & 2 \\ 3 & -1 \end{array}\right]=\left[\begin{array}{rr} 2 & 1 \\ 3 & -2 \end{array}\right] $$

Let $$ A=\left[\begin{array}{rr} 2 & 3 \\ -4 & -5 \end{array}\right] $$ a. Find \(A^{-1}\). b. Show that \(\left(A^{-1}\right)^{-1}=A\).

Let $$ A=\left[\begin{array}{rr} 6 & -4 \\ -4 & 3 \end{array}\right] \text { and } B=\left[\begin{array}{ll} 3 & -5 \\ 4 & -7 \end{array}\right] $$ a. Find \(A B, A^{-1}\), and \(B^{-1}\). b. Show that \((A B)^{-1}=B^{-1} A^{-1}\).

Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, give an example to show why it is false. If \(A^{-1}\) does not exist, then the system \(A X=B\) of \(n\) linear equations in \(n\) unknowns does not have a unique solution.

Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, give an example to show why it is false. If \(A\) is a square matrix with inverse \(A^{-1}\) and \(c\) is a nonzero real number, then $$ (c A)^{-1}=\left(\frac{1}{c}\right) A^{-1} $$
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