Chapter 4: Problem 60
Differentiate with respect to the independent variable. $$ f(x)=\sqrt{x}\left(x^{4}-5 x^{2}\right) $$
Short Answer
Expert verified
The derivative is \( f'(x) = 5x^{3.5} - 6x^{1.5} \).
Step by step solution
01
Apply the Product Rule
To differentiate the function \( f(x) = \sqrt{x}(x^4 - 5x^2) \), recognize that it is a product of \( u = \sqrt{x} \) and \( v = x^4 - 5x^2 \). The product rule states that \( (uv)' = u'v + uv' \).
02
Differentiate \( u = \sqrt{x} \)
To find \( u' \), use the power rule. \( u = x^{1/2} \), so \( u' = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}} \).
03
Differentiate \( v = x^4 - 5x^2 \)
Apply the power rule to each term individually. The derivative of \( x^4 \) is \( 4x^3 \), and the derivative of \( -5x^2 \) is \( -10x \). Therefore, \( v' = 4x^3 - 10x \).
04
Substitute into the Product Rule Formula
Substitute \( u' \), \( v \), and \( v' \) into the product rule formula: \[ f'(x) = \left(\frac{1}{2\sqrt{x}}\right)(x^4 - 5x^2) + (\sqrt{x})(4x^3 - 10x) \].
05
Simplify the Expression
Simplify each term separately. The first term becomes: \[ \frac{x^4 - 5x^2}{2\sqrt{x}} = \frac{x^{7/2} - 5x^{3/2}}{2} \].The second term becomes: \[ 4x^{3.5} - 10x^{1.5} \].Combine to get: \[ f'(x) = \frac{x^{7/2} - 5x^{3/2}}{2} + 4x^{3.5} - 10x^{1.5} \].
06
Finalize the Simplification
Combine the terms for the final expression. After combining like terms, the result is: \[ f'(x) = 5x^{3.5} - 6x^{1.5} \].
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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 a fundamental technique in calculus used to differentiate functions that are products of two or more functions. It offers a solution to the problem of finding the derivative of a product. If you have two functions, say \( u(x) \) and \( v(x) \), their product is \( u(x) \cdot v(x) \). The derivative of this product is given as:\[(uv)' = u'v + uv'\]This formula is crucial as it breaks down the derivative into a manageable form by taking the derivative of one function while keeping the other unchanged and then doing the reverse.
- \( u' \) is the derivative of \( u \)
- \( v' \) is the derivative of \( v \)
Power Rule
The power rule is a simple, yet powerful tool in calculus for differentiating terms of the form \( x^n \), where \( n \) is any real number. According to the power rule, the derivative of \( x^n \) is:\[ \frac{d}{dx} x^n = nx^{n-1} \]This concept is straightforward; simply multiply the term by its power and reduce the power by one. The power rule is evident in the exercise when differentiating both parts of the product.For instance, \( u = x^{1/2} \) is differentiated to become \( u' = \frac{1}{2}x^{-1/2} \), applying the power rule for a root. Similarly, for \( v = x^4 - 5x^2 \), the derivatives are \( v' = 4x^3 - 10x \). It's worth noting how quickly the power rule allows differentiation, even when multiple terms are involved.
- Applicable to power functions
- Fast and efficient differentiation method
Calculus
Calculus, a branch of mathematics, plays a crucial role in solving complex problems involving change and motion. At its core, calculus focuses on two main concepts: differentiation and integration. Differentiation allows us to find how a function changes at any point, giving us the slope of the tangent line to the curve of a function.
- Understanding how variables change
- Applications in physics, engineering, and economics