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# (a) If $f(x) = x \sqrt {2 - x^2},$ find $f'(x).$(b) Check to see that your answer to part (a) is reasonable by comparing the graph of $f$ and $f'$.

## a. $\frac{2-2 x^{2}}{\sqrt{2-x^{2}}}$b. see explanation

Derivatives

Differentiation

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### Video Transcript

all right. Here we have our function f of X equals x times a square root of two minus x squared. We want to find its derivative. So first I'm going to write it as X times two minus X squared to the 1/2 power. And then to find the derivative, we're going to use the product rule. So what we see here is the first times a derivative of a second. And when we take the derivative of the second, we're using the chain rule. So what we see here is a derivative of the 1/2 power part the outside function. We bring down the 1/2 and we raise the inside to the negative 1/2. And then what we see here is the derivative of the insight. The derivative of negative X squared would be negative two X and then moving along. We have the second time's a derivative of the first. The derivative of X is just one. And then we can simplify that derivative a little bit. So what we can do is change our negative 1/2 power back to a radical and put it in the denominator. Since we have a negative power. We multiply the X, the 1/2 and the negative two x and that's where we get negative X squared. All right, So what we want to do for part B of the problem is put our original f of X as well as our f prime into the calculator, look at the graphs and make sure they seem reasonable. So we grab a calculator, we go toe y equals we type f of X into y one. We take we type f prime of X into y two and then for a window. I decided to try negative 3 to 3 on the X axis and negative 626 on the Y axis. And here's what my graphs look like. So let's talk about why these make sense. So I'm going to trace along the graph of the original function F of X now notice as I move from left to right F of X is decreasing. I noticed that its derivative, which is in red, is negative. It makes sense for a derivative to be negative. If a function is decreasing, notice that right about here. The derivative is zero. So the red one is on the X axis. And that's right about where the original function fo backs would have a horizontal tangent line. Now we see that f of X is increasing as we move from left to right. F of X is increasing and we see the derivative. The red one is positive right about here. We see the derivative zero again and we see that is where F of X has a horizontal tangent line. And then now the original is decreasing again and the derivative is negative again, so that all makes sense together.