Problem 17
Question
We showed that the derivative of \(\sqrt{x}\) (or \(x^{\frac{1}{2}}\) ) is \(\frac{1}{2} \frac{1}{\sqrt{x}}\) (or \(\left.\frac{1}{2} x^{-\frac{1}{2}}\right) .\) Here we focus on \(f(x)=\sqrt{x-1}\) (a) How is the graph of \(\sqrt{x-1}\) related to that of \(\sqrt{x}\) ? (b) How is the graph of the derivative of \(\sqrt{x-1}\) related to that of the derivative of \(\sqrt{x} ?\) Illustrate with a rough sketch. (c) Given your answer to part (b) explain why \(\left.\frac{d}{d x} \sqrt{x}\right|_{x=4}=\left.\frac{d}{d x} \sqrt{x-1}\right|_{x=5}\). In other words, explain why the derivative of \(\sqrt{x}\) at \(x=4\) is equal to the derivative of \(\sqrt{x-1}\) evaluated at \(x=5\) (d) Show that \(f^{\prime}(5)=\frac{1}{4}\) using the limit de nition of derivative: $$ f^{\prime}(5)=\lim _{x \rightarrow 5} \frac{f(x)-f(5)}{x-5} $$
Step-by-Step Solution
Verified Answer
a) The function \(\sqrt{x-1}\) is a shift of \(\sqrt{x}\) to the right by 1 unit.\n b) The derivative of \(\sqrt{x-1}\) is a shift of the derivative of \(\sqrt{x}\) to the right by 1 unit.\n c) The derivatives at the points are equal because these points correspond to the same point before and after the shift.\n d) Using the limit definition of the derivative, we find that the derivative of the function at \(x=5\) is \(\frac{1}{4}\).
1Step 1: Relation of \(\sqrt{x-1}\) to \(\sqrt{x}\)
The function \(\sqrt{x-1}\) is a transformation of the function \(\sqrt{x}\). The transformation represents a shift to the right by 1 unit. Thus, every point \((x, y)\) on the graph of \(\sqrt{x}\) turns into the point \((x+1, y)\) on the graph of \(\sqrt{x-1}\).
2Step 2: Relation of the Derivatives
The derivative of \(\sqrt{x}\) given is \(\frac{1}{2\sqrt{x}}\). If we apply the same transformation to the derivative, i.e. replacing \(x\) with \(x-1\), we get the derivative of \(\sqrt{x-1}\) as \(\frac{1}{2\sqrt{x-1}}\)
3Step 3: Explanation of Equal Derivatives at Specific Points
The derivative of a function at a point is equal to the slope of the tangent line at that point on the function's graph. This means that the derivative of \(\sqrt{x}\) at \(x=4\) gives us the slope of the tangent line at that point in the graph of \(\sqrt{x}\). Similarly, the derivative of \(\sqrt{x-1}\) at \(x=5\) gives us the slope of the tangent line at that point in the graph of \(\sqrt{x-1}\). Recall that the graph of \(\sqrt{x-1}\) is obtained by shifting the graph of \(\sqrt{x}\) one unit to the right. Therefore, the tangent line at \(x=4\) on \(\sqrt{x}\) is identical to the tangent line at \(x=5\) on \(\sqrt{x-1}\). Hence, their slopes, i.e., the derivatives at these points, are equal.
4Step 4: Derivative Calculus Using Limit Definition
Using the definition of derivative, \(f^{\prime}(5)=\lim _{x \rightarrow 5} \frac{f(x)-f(5)}{x-5}\), we find: \(f'(5) = \lim _{x \rightarrow 5} \frac{\sqrt{x-1}-2}{x-5}\). Multiply and divide the fraction by its conjugate \(\frac{\sqrt{x-1} + 2}{\sqrt{x-1} + 2}\), we get: \(f'(5) = \lim _{x \rightarrow 5} (\frac{1}{\sqrt{x-1} + 2}) = \frac{1}{4}\)
Key Concepts
Graph TransformationsLimit Definition of DerivativeTangent Line
Graph Transformations
When we talk about graph transformations, we're referring to the ways a graph can be altered by applying certain changes to its equation. One common transformation is shifting the graph.
For example, consider the function \( f(x) = \sqrt{x} \). If we modify it to \( f(x) = \sqrt{x-1} \), we are shifting the entire graph 1 unit to the right. This means every point \((x, y)\) on the graph of \(\sqrt{x}\) moves to \((x+1, y)\) on the graph of \(\sqrt{x-1}\).
This transformation helps in understanding how changes in equations reflect visually. In our exercise, such a shift explains the relationship between \(\sqrt{x}\) and \(\sqrt{x-1}\). It becomes particularly useful when analyzing how the derivative changes between these two functions, as we will see next.
For example, consider the function \( f(x) = \sqrt{x} \). If we modify it to \( f(x) = \sqrt{x-1} \), we are shifting the entire graph 1 unit to the right. This means every point \((x, y)\) on the graph of \(\sqrt{x}\) moves to \((x+1, y)\) on the graph of \(\sqrt{x-1}\).
This transformation helps in understanding how changes in equations reflect visually. In our exercise, such a shift explains the relationship between \(\sqrt{x}\) and \(\sqrt{x-1}\). It becomes particularly useful when analyzing how the derivative changes between these two functions, as we will see next.
Limit Definition of Derivative
The limit definition of a derivative is fundamental in calculus. It tells us how to determine the derivative of a function at a specific point.
The formal definition is \( f'(a) = \lim_{x \rightarrow a} \frac{f(x)-f(a)}{x-a} \). This expression computes the slope of the tangent line to the curve at \(x = a\).
To use this definition properly, let's consider \( f(x) = \sqrt{x-1} \) and find its derivative at \(x = 5\). We use the formula: \( f'(5) = \lim_{x \rightarrow 5} \frac{f(x)-f(5)}{x-5} \).
By substituting \( f(x) = \sqrt{x-1} \), and performing algebraic manipulation like multiplying by the conjugate, we find that the limit evaluates to \( \frac{1}{4} \).
This is a powerful demonstration of using limits to find precise rates of change.
The formal definition is \( f'(a) = \lim_{x \rightarrow a} \frac{f(x)-f(a)}{x-a} \). This expression computes the slope of the tangent line to the curve at \(x = a\).
To use this definition properly, let's consider \( f(x) = \sqrt{x-1} \) and find its derivative at \(x = 5\). We use the formula: \( f'(5) = \lim_{x \rightarrow 5} \frac{f(x)-f(5)}{x-5} \).
By substituting \( f(x) = \sqrt{x-1} \), and performing algebraic manipulation like multiplying by the conjugate, we find that the limit evaluates to \( \frac{1}{4} \).
This is a powerful demonstration of using limits to find precise rates of change.
Tangent Line
A tangent line is a straight line that touches a curve at one point without crossing it. The slope of this line represents the derivative of the function at that specific point.
This is because the tangent line shows the instantaneous rate of change of the function at that location. In practical terms, for a given function like \(\sqrt{x-1}\), the derivative at a point, such as \(x=5\), gives us the slope of the tangent line there.
In our exercise, understanding that \(\sqrt{x}\) shifts to \(\sqrt{x-1}\) explains why the tangent line at \(x=4\) for \(\sqrt{x}\) is the same as at \(x=5\) for \(\sqrt{x-1}\). These points correspond because of the rightward shift of 1 unit.
So, both derivative values (or slopes) are equal, reinforcing the concept of how tangent lines relate to derivatives at given points.
This is because the tangent line shows the instantaneous rate of change of the function at that location. In practical terms, for a given function like \(\sqrt{x-1}\), the derivative at a point, such as \(x=5\), gives us the slope of the tangent line there.
In our exercise, understanding that \(\sqrt{x}\) shifts to \(\sqrt{x-1}\) explains why the tangent line at \(x=4\) for \(\sqrt{x}\) is the same as at \(x=5\) for \(\sqrt{x-1}\). These points correspond because of the rightward shift of 1 unit.
So, both derivative values (or slopes) are equal, reinforcing the concept of how tangent lines relate to derivatives at given points.
Other exercises in this chapter
Problem 16
Find the equation of the line tangent to \(y=\frac{2}{x+1}\) at the point \((1,1)\).
View solution Problem 16
Let \(f(x)=\frac{1}{x^{2}}\). Let \(P\) and \(Q\) be points on the graph of \(f\) with coordinates \((x, f(x))\) and \((x+\Delta x, f(x+\Delta x))\), respective
View solution Problem 18
Let \(g\) be a function that is locally linear. We know that \(g^{\prime}(a)\) is the slope of the tangent line to the graph of \(g\) at point \(A=(a, g(a))\).
View solution Problem 18
Show that \(\frac{d}{d x} \sqrt{x+8}=\frac{1}{2 \sqrt{x+8}}\) using the limit de nition of derivative. You ll use different versions of the de nition in parts (
View solution