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Prove that if \(f(x)=\sqrt{x}\), then \(f^{\prime}(a)=1 /(2 \sqrt{a}),\) for \(a>0 .\) (The expression you obtain for \([f(a+h)-f(a)] / h\) will require some algebraic face lifing. but the answer should suggest the right trick.)

Short Answer

Expert verified
The derivative of \(\sqrt{x}\) is \(\frac{1}{2\sqrt{a}}\).

Step by step solution

01

Write the definition of the derivative

To find the derivative of the function, we start with the definition of the derivative: \[ f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \]
02

Substitute the function into the limit

Since \( f(x) = \sqrt{x} \), substitute \( f(x) \) into the limit definition: \[ f'(a) = \lim_{h \to 0} \frac{\sqrt{a+h} - \sqrt{a}}{h} \]
03

Rationalize the numerator

Next, rationalize the numerator by multiplying by the conjugate. Multiply both the numerator and the denominator by \( \sqrt{a+h} + \sqrt{a} \): \[ f'(a) = \lim_{h \to 0} \frac{(\sqrt{a+h} - \sqrt{a})(\sqrt{a+h} + \sqrt{a})}{h(\sqrt{a+h} + \sqrt{a})} \]
04

Simplify the expression

Simplify the numerator using the difference of squares: \[ (\sqrt{a+h} - \sqrt{a})(\sqrt{a+h} + \sqrt{a}) = (a+h) - a = h \] Thus, the expression becomes: \[ f'(a) = \lim_{h \to 0} \frac{h}{h(\sqrt{a+h} + \sqrt{a})} \]
05

Cancel out the common term

Cancel \( h \) in the numerator and the denominator: \[ f'(a) = \lim_{h \to 0} \frac{1}{\sqrt{a+h} + \sqrt{a}} \]
06

Evaluate the limit

Finally, evaluate the limit as \( h \) approaches 0: \[ \lim_{h \to 0} \frac{1}{\sqrt{a+h} + \sqrt{a}} = \frac{1}{\sqrt{a} + \sqrt{a}} = \frac{1}{2\sqrt{a}} \] So, the derivative is: \[ f'(a) = \frac{1}{2\sqrt{a}} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Definition of the Derivative
The concept of the derivative is central to calculus. It's a way to measure how a function changes as its input changes. For a given function, the derivative at a certain point tells us the slope of the tangent line to the curve at that point.
In mathematical terms, the derivative of a function \( f(x) \) at a point \( a \) is defined as:
\[ f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h} \]
This definition is known as the limit definition of the derivative. It uses the concept of a limit to describe the rate of change of the function as the input changes by an infinitesimally small amount \( h \).
When working with the function \( f(x) = \sqrt{x} \), we start by plugging it into this definition of the derivative.
Limit Process
The limit process is a fundamental step in finding derivatives using the definition. After substituting \( f(x) = \sqrt{x} \) into the derivative definition, we get:
\( f'(a) = \lim_{h \to 0} \frac{\sqrt{a + h} - \sqrt{a}}{h} \)
Here, we need to evaluate what happens to the expression as \( h \) gets very close to 0. This can be tricky because the numerator \( \sqrt{a + h} - \sqrt{a} \) and the denominator \( h \) both tend to 0, leading to an indeterminate form \( \frac{0}{0} \). To resolve this, further algebraic manipulation is required, such as rationalizing the numerator.
Rationalizing the Numerator
Rationalizing the numerator is a useful algebraic technique to simplify expressions involving square roots. In our case, we multiply the numerator and the denominator by the conjugate of the numerator. The conjugate of \( \sqrt{a + h} - \sqrt{a} \) is \( \sqrt{a + h} + \sqrt{a} \). Multiplying both the numerator and the denominator by this conjugate gives:
\( f'(a) = \lim_{h \to 0} \frac{(\sqrt{a + h} - \sqrt{a})(\sqrt{a + h} + \sqrt{a})}{h(\sqrt{a + h} + \sqrt{a})} \)
This manipulation helps us eliminate the square roots in the numerator by turning it into a difference of squares.
Difference of Squares
The difference of squares is a special algebraic formula:
\( (x - y)(x + y) = x^2 - y^2 \)
In our context, applying this formula to \( (\sqrt{a+h} - \sqrt{a})(\sqrt{a+h} + \sqrt{a}) \) gives:

\( (\sqrt{a+h})^2 - (\sqrt{a})^2 \)
Which simplifies to \( a + h - a = h \).
So, our expression now looks like:
\( f'(a) = \lim_{h \to 0} \frac{h}{h(\sqrt{a + h} + \sqrt{a})} \)
We're now in a position to cancel out any common factors.
Evaluating Limits
Once we simplify the expression and find common factors, we progress to evaluate the limit. Cancelling out \( h \) in the numerator and denominator gives:
\( f'(a) = \lim_{h \to 0} \frac{1}{\sqrt{a + h} + \sqrt{a}} \)
As \( h \) approaches 0, \( \sqrt{a + h} \) approaches \( \sqrt{a} \). So the limit becomes:

\( f'(a) = \frac{1}{\sqrt{a} + \sqrt{a}} = \frac{1}{2\sqrt{a}} \)
Therefore, the derivative of the square root function is:
\( f'(a) = \frac{1}{2\sqrt{a}} \)

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Most popular questions from this chapter

(a) Suppose that \(f(a)=g(a)=h(a),\) that \(f(x) \leq g(x) \leq h(x)\) for all \(x\) and that \(f^{\prime}(a)=h^{\prime}(a) .\) Prove that \(g\) is differentiable at \(a,\) and that \(f^{\prime}(a)=g^{\prime}(a)=h^{\prime}(a).\) (Begin with the definition of \(g^{\prime}(a)\).) (b) Show that the conclusion does not follow if we omit the hypothesis \(f(a)=g(a)=h(a).\)

Let \(f(x)=x^{2}\) if \(x\) is rational, and \(f(x)=0\) if \(x\) is irrational. Prove that \(f\) is differentiable at \(0 .\) (Don't be scared by this function. Just write out the definition of \(f^{\prime}(0).\))

For each natural number \(n,\) let \(S_{n}(x)=x^{n} .\) Remembering that \(S_{1}^{\prime}(x)=1,\) \(S_{2}^{\prime}(x)=2 x,\) and \(S_{3}^{\prime}(x)=3 x^{2},\) conjecture a formula for \(S_{n}^{\prime}(x) .\) Prove your conjecture. (The expression \((x+h)^{n}\) may be expanded by the binomial theorem.)

Imagine a road on which the speed limit is specified at every single point. In other words, there is a certain function \(L\) such that the speed limit \(x\) miles from the beginning of the road is \(L(x) .\) Two cars, \(A\) and \(B\), are driving along this road; car \(A\) 's position at time \(t\) is \(a(t),\) and \(\operatorname{car} B^{\prime}\) s is \(b(t).\) (a) What equation expresses the fact that car \(A\) always travels at the speed limit? (The answer is not a' \((t)=L(t)\).) (b) Suppose that \(A\) always goes at the speed limit and that \(B\) 's position at time \(t\) is \(A\) 's position at time \(t-1 .\) Show that \(B\) is also going at the speed limit at all times. (c) Suppose, instead, that \(B\) always stays a constant distance behind A. Under what conditions will \(B\) still always travel at the speed limit?

Find \(f^{\prime}(x)\) and also \(f^{\prime}(x+3)\) in the following cases. Be very methodical, or you will surely slip up somewhere. Consult the answers (after you do the problem, naturally). (i) \(\quad f(x)=(x+3)^{5}.\) (ii) \(\quad f(x+3)=x^{5}.\) (iii) \(\quad f(x+3)=(x+5)^{7}.\)

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