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Magazine Chapter 3: Problem 97
Differentiate. $$f(x)=e^{x / 2} \cdot \sqrt{x-1}$$
Short Answer
Expert verified
\[f'(x) = \frac{1}{2}e^{x/2}(\sqrt{x-1} + \frac{1}{\sqrt{x-1}})\]
Step by step solution
01
Identify the type of function
Recognize that the given function is a product of two functions: \[f(x) = e^{x/2} \text{ and } g(x) = \sqrt{x-1}.\]
02
Apply the Product Rule
The product rule for differentiation states: \[\frac{d}{dx} [u(x) \cdot v(x)] = u'(x)v(x) + u(x) v'(x)\]. Here, let \[u(x) = e^{x/2}\] and \[v(x) = \sqrt{x-1}\].
03
Differentiate \(u(x)\)
Differentiate \[u(x) = e^{x/2}\]. Using the chain rule, \[u'(x) = \frac{d}{dx} \[e^{x/2}\] = \[e^{x/2} \cdot \frac{1}{2} \] = \frac{1}{2}e^{x/2}.\]
04
Differentiate \(v(x)\)
Differentiate \[v(x) = \sqrt{x-1}\]. Using the chain rule again, \[v'(x) = \frac{d}{dx} \sqrt{x-1} = \[\frac{1}{2}(x - 1)^{-1/2} \cdot 1\] = \frac{1}{2\sqrt{x - 1}}.\]
05
Apply the product rule
Using the product rule formula, substitute the differentiated parts: \[f'(x) = u'(x)v(x) + u(x)v'(x)\]. This gives us: \[f'(x) = (\frac{1}{2} e^{x/2})(\sqrt{x-1}) + (e^{x/2})(\frac{1}{2\sqrt{x-1}}).\]
06
Simplify the expression
Combine and simplify the terms: \[\frac{\sqrt{x-1}}{2} \cdot e^{x/2} + \frac{e^{x/2}}{2\sqrt{x-1}} = \frac{1}{2}e^{x/2}(\sqrt{x-1} + \frac{1}{\sqrt{x-1}}).\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Product Rule
The product rule is essential when differentiating functions that are multiplied together. For a product of two functions, the rule is: \[ \frac{d}{dx} [u(x) \cdot v(x)] = u'(x)v(x) + u(x) v'(x) \]. For example, in our exercise, we identified \[ u(x) = e^{x/2} \] and \[ v(x) = \sqrt{x-1} \]. To apply the product rule, we find the derivatives of each function separately (\[ u'(x) \] and \[ v'(x) \]) and then plug these into the formula. The product rule helps us handle more complex differentiation tasks efficiently.
Chain Rule
The chain rule is a technique for differentiating composite functions. If you have a function composed of two others, such as \[ z = f(g(x)) \], the chain rule states: \[ \frac{dz}{dx} = f'(g(x)) \cdot g'(x) \]. In our problem, the chain rule helps to differentiate \[ u(x) = e^{x/2} \] and \[ v(x) = \sqrt{x-1} \]. For \[ e^{x/2} \], which is an outer function of \[ x/2 \], we have \[ f'(x) = e^{x/2} \cdot \frac{1}{2} \]. Similarly, for \[ \sqrt{x-1} \], which is the outer function of \[ x-1 \], the chain rule gives us \[ v'(x) = \frac{1}{2}\sqrt{x-1}^{-1} \cdot 1 \]. Using the chain rule simplifies dealing with nested functions.
Exponential Function
Exponential functions are among the most common uses of differentiation and they often arise in problems involving growth or decay. The general form is \[ a^{u(x)} \], and an important rule in calculus is \[ \frac{d}{dx} [a^{u(x)}] = a^{u(x)} \cdot u'(x) \]. In the differentiation process for \[ e^{x/2} \], we treat \[ e^{x/2} \] as our main function and differentiate the exponent \[ x/2 \] separately to get \[ \frac{1}{2}e^{x/2} \]. This characteristic is especially useful for tackling more complex exponentials.
Radical Function
Radical functions, such as square roots, often appear in calculus. A common form is \[ \sqrt{x} \]. The differentiation rule here involves rewriting the radical in exponent form, such as \[ x^{1/2} \]. In our example, \[ \sqrt{x-1} \] becomes \[ (x-1)^{1/2} \]. The chain rule then helps us manage the inner function \[ x-1 \], leading to \[ \frac{d}{dx} \sqrt{(x-1)} = \frac{1}{2\sqrt{(x-1)}} \]. This transformation simplifies differentiation and works for various degrees of roots.
