5–24. For each of the following composite functions, find an i...

Question

5–24. For each of the following composite functions, find an inner function u=g(x) and an outer function y=f(u) such that y=f(g(x)). Then calculate dy/dx.

y = e^√x

Accepted Answer

Hi everyone, let's take a look at this practice of problem dealing with derivatives. This problem says to identify the inner function U equal to G of X, and the outer function Y is equal to f of U, where the composition of Y is equal to F of G of X, and then find the derivative of Y with respect to X. And we're given the function Y is equal toe raised to the square root of the quantity of X2 + 1. We give 4 possible choices as our answers. For choice A, we have U is equal to x2 + 1. F of U is equal to eu, and the derivative of Y with respect to X is equal to X multiplied by E raised to the square root of the quantity of X2 + 1 in quantity divided by the square root of the quantity of X2 + 1. For choice B, you have U is equal to the square root of the quantity of X2 + 1, F of U is equal to eu, and the derivative of Y with respect to X is equal to X multiplied by E raised to the square root of the quantity of X2 + 1 in quantity divided by the square root of the quantity of X2 + 1. For choice C we have U is equal to the square root of the quantity of X2 + 1, F of U is equal to eu, and the derivative of Y with respect to X is equal toe raised to the square root of the quantity of X2 + 1 in quantity divided by the square root of the quantity of X2 + 1. And for choice D we have U is equal to X2 + 1, F of U is equal to eu, and the derivative of Y with respect to X is equal toe raised to the square root of the quantity of X2 + 1 in quantity, divided by the square root of the quantity of X2 + 1. Now the question asks us to find the inner and outer functions. And so let's look at the inner function first, and our inner function needs to be a function of X. and we're going to choose our inner function U to be the exponent of our exponential. So that means that u is going to be equal to the square root of the quantity of X2 + 1. And so that is our inner function identified. And so now if I replace the square root of X2 + 1 in our function Y, that means Y is going to now be a function of U, and this will be our outer function, and it's going to be equal to E raises to the U. So that is now our outer function identified. So now that we've identified the inner and outer functions, we will now look at calculating the derivative of Y with respect to X. And to do this derivative, we're going to be using the chain rule. So recall for the chain rule that the derivative of Y with respect to X is equal to the derivative of Y with respect to u. Multiplied by the derivative of U with respect to X. And so we're going to use our inner and outer functions to calculate these derivatives. So, the first derivative that we will calculate is the derivative of U with respect to X. This is going to be equal to the derivative with respect to X. Of the square root of the quantity of X2 + 1. And to take this derivative, we need to recall a few rules for our derivatives. First ones can be the power rule, which says that the derivative with respect to X of the quantity of X-rays to the n is equal to n multiplied by X raised to the n minus 1. We'll also be taking a derivative of a constant and so recall that the derivative with respect to X of a constant which we will label as C is going to be equal to 0. So here we actually have to really apply the chain rule even to this quantity. Since I have the quantity of X2 + 1 raised to the 12 power, since the square root of a quantity is the same as raising that quantity to the to the 12 power. So when I take this um Derivative of that quantity I'll get 1/2. Multiplied by the quantity of X2 + 1 in quantity raised to the minus 1/2, and then I still have to take a derivative of X2 + 1. Which in this case turns out to be 2 X. So I apply the power rule 2x2 to give me 2 X, and then I take the derivative of 1, which is a constant which gives me 0. So, the 12 factor and 2 will cancel out, and I can just rewrite this as being equal to x divided by. The square root of the quantity of X2 + 1. So here I've just rewritten the quantity of X2 + 1 and quantity rates to the minus 1/2 as 1 divided by the square root of the quantity of X2 + 1. So now I need to take the derivative of Y with respect to you. And this is going to be equal to the derivative with respect to you. Of the quantity of ease to the U. So here we call your rule for taking derivatives of exponentials, and that is that the derivative with respect to X of the quantity of E to the X is just equal to E X. So we get back the same functions. So this means that our derivative of Y with respect to you is just equal to E raise to the U. So we can now use these results to substitute them back into our formula for the chain rule. So here, We will have now D derivative of Y with respect to X, and that's gonna be equal to. R derivative of Y with respect to you, which is E to the U. However, I want my final answer to be a function of X. So for you, we'll replace it with the square root of the quantity of X2 + 1. So we will have E raised to the square root of the quantity of X2 + 1. And then that is going to get multiplied by. are derivative of you with respect to X. And that quantity was X divided by. The square root of the quantity of X2 + 1. And so now I'm just going to rewrite this result, and so this is going to be equal to x multiplied by E raised to the square root of the quantity of X2 + 1 in quantity, divided by the square root of the quantity of X2 + 1. And so that is the answer for our derivative that we were looking for. And so now, if, uh, now that we've identified both our inner and outer functions and have calculated derivative, I can use our answers and compare them to our answer choices. And 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.

5–24. For each of the following composite functions, find an inner function u=g(x) and an outer function y=f(u) such that y=f(g(x)). Then calculate dy/dx.
y = e^√x

Key Concepts

Composite Functions

Chain Rule

Exponential Functions

Watch next