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Chapter 7: Problem 54

Describe the \(x\) -values at which the function is differentiable. Explain your reasoning. $$ y=x^{2 / 5} $$

Short Answer

Expert verified
The function \(y=x^{2 / 5}\) is differentiable for all real values of \(x\).

Step by step solution

01

Understand the Definitions

Recall that a function \(f(x)\) is differentiable at a particular \(x\) -value if the function's derivative exists at that \(x\) -value. Essentially, this means that the function has a defined slope at that point.
02

Differentiate the Function

We need to calculate the derivative of the function. In this case, the function \(y=x^{2 / 5}\) can be differentiated using the power rule, which states that the derivative of \(x^n\) is \(nx^{n?1}\). Applying this rule to the function, we compute the derivative as \(y'=(2 / 5)x^{-3 / 5}\).
03

Determine Where the Derivative Is Undefined

A function is only differentiable where its derivative is defined. For the derivative function \(y'=(2 / 5)x^{-3 / 5}\), it is undefined when the denominator is zero. Since there are no values of \(x\) that will make the denominator zero, the derivative is defined for all real values of \(x\).

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Most popular questions from this chapter

Given \(f(x)=x+1\), which function would most likely represent a demand function? Explain your reasoning. Use a graphing utility to graph each function, and use each graph as part of your explanation. (a) \(p=f(x)\) (b) \(p=x f(x)\) (c) \(p=-f(x)+5\)

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