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# Find the derivative of the function.$y = \sqrt {1 + xe^{-2x}}$

## $\frac{1-2 x}{2 e^{2 x} \sqrt{1+x e^{-2 x}}}$

Derivatives

Differentiation

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### Video Transcript

we're going to use the chain rule to find the derivative of this function. But first I'm going to rewrite it, because when I see a square root, I like to rewrite it as a 1/2 power. That makes it easier to differentiate. Okay, now let's find the derivative. So the outside function is the 1/2 power function. So we bring down the 1/2. We raised the inside to the negative 1/2. Then we multiply by the derivative of the inside and notice that the inside has some. But the derivative of one is just zero. So we're gonna focus on the derivative of the second term, which is a product. So for that, we're going to need to use the product rule. The first is X, and the second is e to the negative two x So the first x times, the derivative of the second and the derivative of each of the negative two x would be e to the negative two x times negative too. So that's the first times a second or first times derivative. The second plus the second e to the negative two x times the derivative of the first and the derivative of X is just one. Okay, now let's see what we can do to make this a little bit simpler. So what we can dio is anything that has a negative exponents like this. Here, we can move it to the bottom and we can move the two to the bottom and then we'll clean up this last part a little bit. So we have negative two x e to the negative two x plus e to the negative two x over two times the square root of one plus x e to the negative two x. When I moved the 1/2 power part down to the bottom, I just changed it back to a square root sign. Now, another thing we could do is we could factor e to the negative two x out of both terms from the numerator, so that gives us e to the negative two x times negative two x plus one over two times the square root of one plus x e to the negative two x Now. Earlier, we took things that had negative power and moved him to the bottom. So we should do the same with this so we have now. As long as we're rewriting, we could right this second factor as one minus two x its with one minus two x over to e to the two x times the square root of one plus x e to the negative two x.

Oregon State University

Derivatives

Differentiation

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