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If $ g(x) = \sqrt {f(x)}, where Ihe graph of $ f $ is shown, evaluate $ g'(3). $

$-\frac{\sqrt{2}}{6}$

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All right. So here we have a graph of F, and you should definitely use the graph shown in your book and not rely on line because mine's just a rough sketch and it might not be quite as accurate. Our goal is to find g prime of three, and we know that G of X is the square root of f of X. I'm going to rewrite G of X as f of X to the 1/2 power, and then we're going to differentiate it using the chain rule. The outer function is the 1/2 power function. So we bring down the 1/2 and we raise the inside to the negative 1/2 power. Then we multiply by the derivative of the inside, and that would be f prime of X. So now we want to find your prime of three. So we substituted three. And for X, it's we've 1/2 times f of three to the negative 1/2 power, tens of crime of three. So we can use the graph to find f of three. So when X equals three, it looks like the graph is at a height of about two. So f of three is too. So now we have 1/2 times to to the negative 1/2 power. That would mean one over square root, too. Now we can also use the graph to find f prime of three. F Prime would be the derivative. So that's the slope of the tangent line and noticed that I have drawn and read the tangent line to the Point X equals three, and it looks like that line has a slope of negative 2/3. You can see the rise of two for the run of three, so we're going to substitute negative 2/3 in there for F prime of three. Now we just need to simplify. And so we have a two on the top of the two on the bottom. We can cancel those, and we have. We're changing our two to the negative 1/2 power and writing it as one over square root, too. So we have one over square root, too. Times negative 1/3. Now, if we don't like to leave a radical in the denominator than we need to take this and multiply it by special form of one square to over square or two, and that's going to give us negative square root, too. Over six

Oregon State University