One-Sided Derivatives

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Question

One-Sided Derivatives

Compute the right-hand and left-hand derivatives as limits to show that the functions in Exercises 37–40 are not differentiable at the point P.

Graph showing a point P(1,1) on the curve y=sqrt(x) and a tangent line y=2x-1, illustrating differentiability concepts.

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says determine whether the function is differentiable at the given point by evaluating the right hand and left hand derivatives using limits. Below the problem we're given a graph that has our function plotted on it, and our function consists of two pieces. The first piece goes between X equal to 0 and X equal to 4. It is a curved function, and it follows the curve Y is equal to square root of X. The second piece is a linear piece, and it begins at X equal to 4 and ends at X equals to 6. And it follows the equation, Y is equal to 2 X minus 6. We're also given the point Q to look at, and that point Q is located at 4.2. We're also given two possible choices as our answers. For choice A, we have yes, and for choice B, we have no. Now, we need to determine whether the function is differentable at the point Q indicated in the graph by evaluating the right hand and left hand derivatives using limits. So, we're gonna start off by looking at the left hand derivative. And using the limit formulation, recall that this is going to be the limit. As x approaches or from the left. Of the quantity of F of X. Minus F of 4 in quantity, divided by the quantity of X minus 4. Now, since X is going to be slightly less than 4, we're going to want to use the square root of X 4 F of X. So this is going to be equal to the limit, as X approaches 4 from the left of the quantity of the square root of X. Minus the square root of 4 in quantity, divided by the quantity of X minus 4. So simplifying this expression, this is going to be equal to the limit, as X approaches 4 from the left. Of the quantity of the square root of X minus 2 in quantity, divided by. The quantity of X minus 4 in quantity. So we want to remove the radical out of the numerator. So we can multiply both the numerator and denominator by the conjugate of the numerator. So we want to multiply this by the square root of X. Plus 2, and that quantity is going to be divided by the quantity of the square root of X plus 2. So multiplying everything out, this is going to be equal to the limit, as X approaches 4 from the left. Of the quantity of X minus 4 in quantity, divided by the quantity of X minus 4 in quantity multiplied by the quantity of the square root of X plus 2 in quantity. The factor of X minus 4 will cancel out both the numerator and denominator, and when we're left with, it's going to be equal to the limit. As x approaches or from the left of the quantity of 1 divided by the quantity of the square root of X plus 2 in quantity. We can evaluate this limit by direct substitution, and so this is going to be equal to 1 divided by the quantity of the square root of 4, plus 2 in quantity, which is equal to 1 divided by 4. So now we're going to evaluate our right hand derivative. And we call for the right-hand derivative that is going to be the limit as X approaches or from the right of the quantity of F of X minus F of 4 in quantity, divided by the quantity of X minus 4 in quantity. And so, here, X is going to be slightly greater than 4, so we're going to want to use 2 X minus 6 as F of X. So this is going to be equal to the limit. As X approaches 4 from the right. Of the quantity of 2 X minus 6. Minus the quantity of 2 multiplied by 4. 6 in quantity, and all of that is divided by the quantity of X minus 4. So simplifying the numerator, this is going to be equal to the limit, as X approaches 4 from the right. Of the quantity Of 2 X. Minus 8 in quantity, divided by the quantity of X minus 4. Now I can factor out the 2 from each of the terms in the numerator, and so doing so, this is going to be equal to the limit, as X approaches 4 from the right. Of the quantity of 2 multiplied by the quantity of X minus 4 in quantity, divided by the quantity of X minus 4. Again, the factor of X minus 4 will cancel out, and what I'm left with is the limit, as X approaches 4 from the right of 2. So, here I'm taking the limit of a constant, so I'm just going to get back that constant. So this is going to be equal to 2. And here I see that my left hand derivative is not equal to my right hand derivative, since 1 divided by 4 is not equal to 2. And since the left-hand derivative is not equal to the right hand derivative. That means our function is not differentiable at the point that we have on the graph. So, the answer to this problem is actually no. And if we compare our answer here of no to our answer choices, we see that the correct answer choice corresponds to answer choice B. So, I hope that this explanation has been useful, and I'll see you in the next video.

One-Sided Derivatives

Compute the right-hand and left-hand derivatives as limits to show that the functions in Exercises 37–40 are not differentiable at the point P.

Key Concepts

One-Sided Derivatives

Differentiability

Limits

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