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Chapter 3: Problem 42

A derivative formula a. Use the definition of the derivative to determine \(\frac{d}{d x}(\sqrt{a x+b}),\) where \(a\) and \(b\) are constants. b. Let \(f(x)=\sqrt{5 x+9}\) and use part (a) to find \(f^{\prime}(x)\) c. Use part (b) to find \(f^{\prime}(-1)\)

Short Answer

Expert verified
Answer: \(\frac{d}{d x}(\sqrt{ax+b}) = \frac{a}{2\sqrt{ax+b}}\) b. Find the derivative of the function \(f(x) = \sqrt{5x + 9}\) using the result from part (a). Answer: \(f'(x) = \frac{5}{2\sqrt{5x+9}}\) c. Calculate the value of \(f'(x)\) at \(x = -1\) for the function in part (b). Answer: \(f'(-1) = \frac{5}{4}\)

Step by step solution

01

Use the definition of the derivative for part (a)

First, recall the definition of the derivative: $$\frac{d}{d x}f(x) = \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}$$ Our function is \(f(x) = \sqrt{ax + b}\), where \(a\) and \(b\) are constants. Apply the definition of the derivative to this function: $$\frac{d}{d x}(\sqrt{ax+b})=\lim_{h \to 0} \frac{\sqrt{a(x+h)+b}-\sqrt{ax+b}}{h}$$
02

Rationalize the numerator

To simplify the expression, multiply the numerator and denominator by the conjugate of the numerator, which is \(\sqrt{a(x+h)+b}+\sqrt{ax+b}\): $$\frac{d}{d x}(\sqrt{ax+b}) = \lim_{h \to 0}\frac{(\sqrt{a(x+h)+b}-\sqrt{ax+b})(\sqrt{a(x+h)+b}+\sqrt{ax+b})}{h(\sqrt{a(x+h)+b}+\sqrt{ax+b})}$$ This will simplify the numerator: $$\frac{d}{d x}(\sqrt{ax+b})=\lim_{h \to 0}\frac{a(x+h)+b-(ax+b)}{h(\sqrt{a(x+h)+b}+\sqrt{ax+b})}$$
03

Simplify the expression and find the limit

Further simplification yields: $$\frac{d}{d x}(\sqrt{ax+b})=\lim_{h \to 0}\frac{ah}{h(\sqrt{a(x+h)+b}+\sqrt{ax+b})}$$ Now, cancel out the \(h\) term by dividing both the numerator and denominator: $$\frac{d}{d x}(\sqrt{ax+b})=\lim_{h \to 0}\frac{a}{\sqrt{a(x+h)+b}+\sqrt{ax+b}}$$ Finally, find the limit as \(h\) approaches 0: $$\frac{d}{d x}(\sqrt{ax+b})=\frac{a}{\sqrt{ax+b}+\sqrt{ax+b}}=\frac{a}{2\sqrt{ax+b}}$$
04

Use the result from part (a) to find \(f'(x)\) for the function in part (b)

For part (b), \(f(x) = \sqrt{5x+9}\), so \(a = 5\) and \(b = 9\). Applying the result from part (a): $$f'(x) = \frac{a}{2\sqrt{ax+b}}=\frac{5}{2\sqrt{5x+9}}$$
05

Calculate the value of \(f'(x)\) at the given point for part (c)

Now that we have the derivative, we need to find the value of \(f'(x)\) at \(x = -1\). Simply plug in \(x = -1\) into the derivative function: $$f'(-1) = \frac{5}{2\sqrt{5(-1)+9}}=\frac{5}{2\sqrt{4}}=\frac{5}{4}$$ The values of the derivative for each part of the problem are as follows: a. \(\frac{d}{d x}(\sqrt{ax+b}) = \frac{a}{2\sqrt{ax+b}}\) b. \(f'(x) = \frac{5}{2\sqrt{5x+9}}\) c. \(f'(-1) = \frac{5}{4}\)

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