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Problem 98 Hard Difficulty

(a) Write $ \mid x \mid = \sqrt {x^2} $ and use the Chain Rule to show that
$ \frac {d}{dx} \mid x \mid = \frac {x}{\mid x \mid} $
(b) If $ f(x) = \mid \sin x \mid, $ find $ f'(x) $ and sketch the graph of $ f $ and $ f'. $ Where is $ f $ is not differentiable?
(c) If $ g(x) = \sin \mid x \mid, $ find $ g'(x) $ and sketch the graphs of $ g $ and $ g'. $ Where is $ g $ not differentiable?

Answer

a. $\frac{1}{2 \sqrt{u}} \cdot u^{\prime}$
b. $f^{\prime}(x)=\frac{\sin 2 x}{2|\sin x|}$
$f$ is not differentiable when $x=n \pi$
c. $g^{\prime}(x)=\left\{\begin{array}{ll}{\cos x} & {\text { if } x \geq 0} \\ {-\cos x} & {\text { if } x < 0}\end{array}\right.$ $f$ is not differentiable at $x=0$

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Video Transcript

In this problem, we have the statement that the absolute value of X is equal to the square root of X squared. So let's use that idea to find the derivative of the absolute value of X, so that would be equal to the derivative of the square root of X squared. And we can use the chain rule on that. So the square root function is the outside function, and it's derivative. If we think of it as a 1/2 power would be 1/2 times the inside to the negative 1/2. And then we multiply by the derivative of the insight. We have the derivative of the outside times, the derivative of the inside, and we can simplify that. We can multiply the 1/2 and the two x together, and that gives us X. And then we can take the part that's to the negative 1/2 power. Bring it down to the denominator and change back to a square root. So then, if we substitute the absolute value of X in four the square root of X squared, we end up with X over the absolute value of X. That's the derivative of the absolute value of X, and that was what we set out to show. Now we can apply that to a couple of problems. So we have part B. Where are function is the absolute value of the sign of X. So it's not just the absolute value of X, it's the absolute value of the sign of X. So we're going to be using the chain rule along with this derivative that we just found. So what Will Dio is a derivative of the outside would be the function over the absolute value of a function multiplied by the derivative of the inside. So the derivative of Sinus coastline. Okay, now we're going to graph that we're going to graph the function and it's derivative and see what we think of what we see. So we go to the calculator, we go toe y equals and we type in our functions. F of X equals the absolute value of sign of X. And then it's derivative. And for a window I chose to go from oh, roughly negative two pi or so a little bit more. Maybe around negative three pie Somewhere in there doesn't matter. We could change these we could make it negative. Two pi to two pi and then the Y values Negative 3 to 3 and we compress graph. Okay, so first of all, let's turn off the derivative function. Unless just look at the original function. Do you see why that would represent the absolute value of the sine function? Sine function normally has this curve, but then it normally goes down here. However, because we took the absolute value, it flipped all the negative parts up to the top. Anything that was below the X axis, it flipped it upside down. So we just have all of these curves above the X axis. And so for the derivatives, let's turn that derivative equation back on notice. We have all these separate segments. The derivative is not defined at zero. It's not defined at pie. It's not defined it to pie at negative pie at negative two pie. It's not defined at any multiple of pie. So will say it's not defined at in times pi, where in is an integer. Now let's go ahead and do part C. So we're back to finding the derivative of a function, and this time our function is G of X and it's the sign of the absolute value of X. So we're using the same two functions. But in a different order. We're still going to need to use the chain rule to find the derivative. So the derivative of Sign is co sign. So the derivative of the outside function is the co sign of absolute value, Becks. And then we multiply by the derivative of the inside function. And that's where we get the X over the absolute value of X. So let's go ahead and take a look at the graph of that as well. Alright, So back to the calculator and back to y equals. So here we have typed in the original function G of X and it's derivative G prime of X. Now I only have the original function. Why one turned on for the moment. So let's take a look at its graph. So what happened was we took the coastline graph actually a sign graph. Excuse me and we took the right side and it ended up reflected across to the left side because we took the absolute value of X. So any X value that was negative ends up having the same Y value as the corresponding X value. That was positive. Okay, so then let's turn on the derivative graph and it's the red one. So it's defined everywhere. Accepted X equals zero. And that's because if you think back about what we had, we had a denominator of absolute value backs and right over here and so we can't divide by zero, everything else was to find, so there we have it.