Find the following higher-order derivatives.
dⁿ

Question

Find the following higher-order derivatives.

dⁿ/dxⁿ (2^x)

Accepted Answer

Hi everyone. Let's take a look at this practice of problem dealing with derivatives. This problem says to find the nth derivative of Y is equals 214 rates to the power X. We're given 4 possible choices as our answers. For choice A, we have N multiplied by the natural log of 14. For choice B, we have 14 rates to the power X, multiplied by the quantity of the natural log of 14, and quantity rates to the power n. For choice C, we have the quantity of the natural log of 14 and quantity rates to the power X, and for choice D, we have 14 rates to the power X, multiplied by the quantity of N multiplied by the natural log of 14 and quantity rates to the power N. Now we're asked to find the derivative of Y for this problem. So we're gonna start off by taking derivatives of Y. So we'll start off by taking the first derivative of Y. With respect to X, we'll have the derivative of Y with respect to X, and this is going to be equal to the derivative with respect to X. Of 14 raise to the power X. So here we need to recall our rules that the derivative with respect to X, Of the quantity of B raised to the power X where B here is a constant is going to be equal to B raised to the power X multiplied by the natural log of B. So apply in this rule to our derivative, this is going to be equal to 14 raised to the power X multiplied by the natural log of 14. So that is our first derivative. Let's take the second derivative and see if we can notice a pattern to help us determine the correct answer. So, we'll have the second derivative of Y with respect to X. And this is going to be equal to the derivative with respect to X of our first derivative, which is the quantity of 14 raised to the power X, multiplied by the natural log of 14. Now, the natural log of 14 is a constant, so I can pull it through the derivative. So we'll have the natural log of 14. That gets multiplied by the derivative spec X of 14 rates to the power X, and we actually calculated that, that is basically our first derivative again. So we'll multiply the natural log of 14 by the quantity of 14, rated to the power X multiplied by the natural log of 14. So simplifying this expression, this is going to be equal to 14 rates to the power X, multiplied by the quantity of the natural log of 14 in quantity squared. So let's calculate the 3rd derivative now. We'll have the 3rd derivative of Y with respect to X. This is going to be equal to the derivative with respect to X of our second derivative, which is the quantity of 14 raised to the power X, multiplied by the quantity of the natural log of 14 in quantity squared. So again, taking this derivative, the quantity of the natural log of 14 and quantity squared is a constant, so I can pull it out in front of the derivative. So this is going to be equal to the quantity, the natural log of 14 in quantity squared. And then I'll be taking the derivative with respect to X of 14 rates to the power X, which we've already seen is 14 rated to the power X multiplied by the natural log of 14. To simplify this expression, this is going to be equal to 14, raise to the power X, multiplied by the quantity of the natural log of 14, in quantity cued. So now we can start to notice a pattern that every time I take a derivative, I'm going to be multiplying our quantity by the natural log of 14. So when I took the first derivative, we had 14 rates to the power X multiplied by the natural log of 14. When we took a second derivative, we have 14 rates to the power X multiplied by the quantity of the natural log of 14 and quantity squared. And when we took our third derivative, we had to multiply our 14 rates to the power X by 3 factors of the natural log of 14, so we had the quantity of the natural log of 14 cubed, multiplying that 14 rates to the power X. So that means that every time we take a derivative, we multiply 14 rates to the power X by the natural log of 14. So, if I'm looking for the nth derivative of Y with respect to X, That is going to be equal to 14, raised to the power X, but it's gonna have to multiply that by the natural log of 14 in times. So that 14 rates to the power X gets multiplied by the natural log of 14, and that quantity is going to be raised to the power N. So, this is the answer that we're actually looking for. And if we compare our answer here 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.

Find the following higher-order derivatives.
dⁿ/dxⁿ (2^x)

Key Concepts

Higher-Order Derivatives

Exponential Functions

Leibniz Notation

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