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In Exercises \(29-32,\) find \(d y / d x\) $$y=2 \sqrt{x}-\frac{1}{\sqrt{x}}$$

Short Answer

Expert verified
The derivative of the function \(y = 2\sqrt{x} - 1/\sqrt{x}\) is \(dy/dx = x^{-1/2} + x^{-3/2}\).

Step by step solution

01

Rewrite the function into a form that is easier to differentiate

Express the square root functions as fractional exponents. Rewrite \(y = 2\sqrt{x} - 1/\sqrt{x}\) as \(y = 2x^{1/2} - x^{-1/2}\).
02

Apply the power rule

Apply the power rule of differentiation, which states that the derivative of \(x^n\) is \(nx^{n-1}\). The derivative of \(2x^{1/2}\) is \(2*(1/2)*x^{1/2 - 1} = x^{-1/2}\), and the derivative of \(-x^{-1/2}\) is \(-(-1/2)*x^{-1/2 - 1} = x^{-3/2}\).
03

Simplify the derivative

Simplify the resulting expression for the derivative. In this case, though, the expression is already fully simplified, so no further action is required. Therefore, the derivative of \(y = 2x^{1/2} - x^{-1/2}\) is \(dy/dx = x^{-1/2} + x^{-3/2}\).

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Key Concepts

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

Power Rule
When differentiating radical functions, one of the most fundamental tools is the power rule for derivatives. This rule asserts that the derivative of a function in the form of a monomial \( x^n \) with respect to x is \( nx^{n-1} \). To put it simply, we multiply the function by the power and then subtract one from the power.For differentiation of radical functions, the radicals are first converted into fractional exponents, as this allows for direct application of the power rule. For example, differentiating \( x^{1/2} \) using the power rule gives us \( \frac{1}{2}x^{1/2-1} = \frac{1}{2}x^{-1/2} \), demonstrating how the exponent is decreased by one. Understanding and correctly applying the power rule is crucial to finding the derivative of any polynomial or radical function, making it an essential technique in calculus.
Fractional Exponents
Fractional exponents are a pivotal concept in calculus, particularly when dealing with radical functions. Expressing roots as fractional exponents can make the process of differentiation more streamlined and cohesive. The nth root of x can be equivalently written as \( x^{1/n} \).In the example \( y = 2\sqrt{x} - \frac{1}{\sqrt{x}} \), the square roots are converted into fractional exponents to become \( y = 2x^{1/2} - x^{-1/2} \). This transformation is not just for stylistic purposes; it equips us to use the power rule for differentiation directly, as seen in the next steps. For students encountering derivatives for the first time, mastering this conversion from root to fractional exponent is an essential skill and will serve as a strong foundation for more complex calculus problems.
Derivative Simplification
Once the power rule has been applied, the next part is the simplification of the derivative. Simplification involves rearranging and combining like terms to present the derivative in its most reduced form. For the function \( y = 2x^{1/2} - x^{-1/2} \), after applying the power rule, we get the derivative \( dy/dx = x^{-1/2} + x^{-3/2} \). While this expression is already simple, it shows the pattern of exponents decreasing by one unit.Simplifications can also include rationalizing denominators, factoring, and canceling out terms. However, it's important not to over-simplify and lose critical information about the function's behavior. Efficiently simplifying derivatives can make further analysis, such as interpreting the behavior of the original function or finding critical points, much clearer for students.

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Most popular questions from this chapter

Find an equation for a line that is normal to the graph of \(y=x e^{x}\)and goes through the origin

Multiple Choice Which of the following is equal to the slope of the tangent to \(y^{2}-x^{2}=1\) at \((1, \sqrt{2}) ?\) ? \((\mathbf{A})-\frac{1}{\sqrt{2}} \quad(\mathbf{B})-\sqrt{2}\) \((\mathbf{C}) \frac{1}{\sqrt{2}} \quad(\mathbf{D}) \sqrt{2} \quad(\mathbf{E}) 0\)

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In Exercises \(1-28\) , find \(d y / d x\) . Remember that you can use NDER to support your computations. $$y=\log _{10} e^{x}$$

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