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 = √x²+1

Accepted Answer

Hi everyone, let's take a look at this practice 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 for the composition Y is equal to F of GFX, and then to find the derivative of Y with respect to X, and we're given the function Y is equal to the square root of the quantity of XQ plus 126. We're given 4 possible choices as our answers. For choice A, we have U is equal to xq plus 126. F of U is equal to. You raised to the 1/2, and the derivative of Y with respect to X is equal to minus 3 x 2 divided by the quantity of 2 multiplied by the square root of the quantity of XQ plus 126. For choice B, we have U is equal to XQ plus 126. F of U is equal to u raised to the 12, and the derivative of Y with respect to X is equal to 3x 2, divided by the quantity of 2 multiplied by the square root of the quantity of X cubed plus 126. For choice C, we have U is equal to x raises to the 1/2, F of U is equal to uq plus 126, and the derivative of Y with respect to X is equal to 3X 2 divided by the quantity of 2 multiplied by the square root of the quantity of XQ plus 126. And finally for choice D, we have U is equal to XQ plus 126. F of U is equal to UQ plus 126, and the derivative of Y with respect to X is equal to 3X 2 divided by the square root of the quantity of XQ plus 126. Now we need to begin by identifying the inner and outer functions. Given the function that Y is equal to the square root of the quantity of XQ plus 126, and so our inner function needs to be a function of X. And so if we look at the function that we're given for Y, we're going to choose our inner function to be what is underneath the radical. That means that you. It's going to be equal to X cubed plus 126. And so that is going to be our inner function. So we can use this inner function to write Y now, which is going to be a function of you. And so, rewriting Y in terms of you, that means that Y is going to be equal to you raised to the 1/2 power. So recall that a square root is raising the quantity to the power of 12. And so this is going to be our outer function. So now we have the inner and outer functions identified. So the next thing we need to calculate is the derivative of Y with respect to X. And to do this, we're going to use the chain rule. So recall for the chain rule that will have the derivative of Y with respect to X, and that's going to be equal to the derivative of Y with respect to U, multiplied by the derivative of U with respect to X. So we're going to use our inner and outer functions to calculate these derivatives. So we're first going to calculate the derivative of U with respect to X. So that is going to be equal to the derivative with respect to X of the quantity of X cubed plus 126. And when I take the derivative of this function, this is going to be equal to 3 x 2. So that recall that if I have a variable raised to a power, and I take a derivative with respect to that variable, that I multiply that variable by the exponent and then subtract one from the exponent. So here in this case, I've multiplied the X by 3 and subtracted 1 from 3 to get an exponent of 2. Now we need to look at 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 you raised to the one half power. And when I take the derivative of this function, this is going to be equal to 1/2 multiplied by U raised to the minus 1/2 power. So again, here, we've applied the same rule as we did before, where we multiply the variable by the exponent, in this case 1/2, and then subtract one from the exponent. So if I have 1/2 minus 1, that gives me minus 12. So now I have these two quantities calculated, I can substitute in those expressions back into our formula for the chain rule. So we'll have the derivative of Y with respect to X, and that could be equal to the derivative of Y with respect tou, which is going to be 1/2, multiplying the quantity, and here for you, I'll substitute back in for you, X cubed plus 126, because we want our final answer as a function of X. So I'll have 1/2 multiplying the quantity of X cubed. Plus 126 in 20 raised to the minus 12 power. And then I need to multiply this by our derivative of you with respect to X, which we calculated to be 3 X squared. And so we can simplify this expression a bit. We'll have the derivative of Y, with respect to X. It's equal to 3 x squared Divided by the quantity of 2 multiplied by the square root of the quantity of X cubed plus 126. So here I have rewritten the quantity of X cubed plus 126 in quantity raised to the minus 1/2 as 1 divided by the square root of the quantity of XQ plus 126. So this is the answer that I was looking for for the derivative. So now that I have both our inner and outer functions identified, as well as the the derivative of Y with respect to X calculated, I can look at my 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 = √x²+1

Key Concepts

Composite Functions

Chain Rule

Differentiation of Square Roots

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