Problem 54 - Calculus an Applied Approach

Question

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

Step-by-Step Solution

Answer

The function \(y=x^{2 / 5}\) is differentiable for all real values of x except \(x=0\).

Step 1: Differentiate the function

The derivative of the function can be found using the power rule. The power rule tells us that the derivative of \(x^n\) is \(n\cdot x^{n - 1}\). Here, \(n=2/5\). So the derivative of \(y=x^{2 / 5}\) would be \((2/5)\cdot x^{(2 / 5) - 1}\). Carry out the subtraction in the exponent yields: \((2/5)\cdot x^{-3 / 5}.\)

Step 2: Identify where the derivative is undefined

The derivative \((2/5)\cdot x^{-3 / 5}\) will not be defined when \(x=0\) because it results in dividing by zero situation which is undefined in mathematics. Therefore, the derivative is not defined at \(x=0\). This implies that the function \(y=x^{2 / 5}\) is not differentiable at \(x=0\).

Step 3: Check for continuity of the derivative

The derivative \((2/5)\cdot x^{-3 / 5}\) is continuous for all \(x \neq 0\), since it is a polynomial function excluding \(x=0\), and polynomial functions are always continuous. Therefore, the function \(y=x^{2 / 5}\) is differentiable for all real values of x, except \(x=0\).

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that involves calculating the derivative of a function. The derivative represents the rate at which the function's output changes with respect to a change in the input. Essentially, if you have a function that describes a certain quantity over time, differentiation tells you how that quantity is changing at any given moment.

For the function in our exercise, the differentiation problem asks which values of the independent variable, in this case, \( x \), allow the function to have a derivative. To perform this operation, we apply specific rules of differentiation that enable us to calculate the gradient of the tangent to the curve at any point on the graph of the function. When a function is not differentiable at a certain point, it may be due to a sharp corner, a vertical tangent, or, as with our function \( y=x^{2/5} \), where the input causes an undefined operation in the derivative formula.

Power Rule

The power rule is a quick and powerful tool for differentiating functions where the variable is raised to a power. According to the power rule, when you have a function in the form of \( x^n \) where \( n \) is any real number, its derivative is \( n \cdot x^{n-1} \). This simplifies the differentiation process significantly.

In the context of our exercise, we used the power rule to differentiate \( y=x^{2/5}\). We applied the rule directly to find the derivative, which became \( (2/5)\cdot x^{(2/5)-1} \). This reduces to \( (2/5)\cdot x^{-3/5}\), which then helps us understand the behavior of the function's slope at different values of \( x \). However, this only works when \( x \) does not create undefined expressions in the derivative, as is the case with \( x = 0 \) in our exercise.

Continuity of Derivatives

The continuity of the derivative of a function is an insightful aspect as it reflects the smoothness of the graph of the function. If the derivative of a function exists and is continuous at a point, the function itself is said to be differentiable at that point. If the derivative is not continuous, the graph of the function may have breaks, cusps, or vertical tangents at the respective points.

In our step-by-step solution, we identified that the derivative \( (2/5)\cdot x^{-3/5}\) is continuous for all \( x \) values except when \( x=0 \). The reason is that, at \( x=0 \) the expression becomes a division by zero, which is undefined in mathematics. Elsewhere, the function behaves nicely – it has no interruptions and the graph of the derivative would be a smooth curve. Therefore, the original function will be differentiable everywhere except at \( x=0 \), maintaining continuity in its derivatives across the other points in its domain.