 Hello there, and welcome to a video where we're going to do some very basic examples of taking derivatives where we have an implicitly defined function involved. We're going to look at more advanced examples in the next video, but this should set the stage for what we need to know here. So we're going to be working with some derivatives where we have x and some function f of x mixed up together. And we don't know what f of x is. We only know that it is a differentiable function of x. So in the reading, sometimes this f of x is referred to as y, but let's go ahead and call it f of x just to emphasize that this is a function that depends upon x in some way about which we don't have complete information. So with that minimal amount of information, let's try to find some derivatives here. So first of all, let's take the derivative of x cubed plus f of x. Well, I'm going to proceed through the derivative taking process just like I normally would. And that means I can use the sum rule to turn this into the derivative with respect to x of x cubed plus the derivative with respect to x of this mystery function f of x. Okay, so part of this expression you see here now is easy and namely this part. This part is just a straight derivative that we were used to from a long time ago and that derivative is just 3x squared. Plus now I have to think about what to do with this thing. I don't actually know what f of x is. I don't have a formula for it. I don't have a graph for it. I only know that it is a differentiable function of x. And so when I'm asked to find out what its derivative is, I can't say much. The only thing I can say is that the derivative of this function is f prime of x. And so that is as far as we can take this derivative here. The derivative process proceeds just like you normally would until we run into this function f or sometimes again in your examples in the reading it's called y. And in that case we just treat the function as a differentiable function and try to write out whatever we can formula wise for its derivative and this is as far as we can go in this particular case here. Now let's look at a second example where we'll apply the same principle. We're taking the derivative of square root of x times f of x. Now this is a product of two things. A radical function times the function f. And so that means I'm going to need to use the product rule on this. And remember that radical x is the same thing as x to the one half for derivative taking purposes. So I'm going to use the product rule to write this out. I would need to take the derivative with respect to x of the first function and multiply the result times the second. Plus take the first function and multiply it times the derivative of the second. That hasn't changed. That is the same product rule we have been using for quite a bit now. What is going to change I think is going to be right here. This is the only thing that's really different between this problem and other product rule problems you've seen in the past. The derivative of radical x is one half x to the minus one half. And I'm multiplying this times f. I don't know what f is so this is as much as I can write. Plus radical x. And now again since I don't know much about f the only thing I can say about its derivative is that it's equal to f prime. And that's my result. It seems like we haven't worked hard enough on this but actually we've done as much as we can with what we have. And so we have this final result here up to some basic algebra simplifications. This is our derivative. One last example here is let's take the derivative of one over f of x and see what we get there. Now just to remind us this is the same thing as taking the function f of x and raising it to the negative one power. So this is really a composite function here. The inside function is f of x and the outside function is the function x to the negative one. And so I'm going to use the chain rule which means I need to take this thing's derivative the outside function that's negative x or whatever to the minus two power and then evaluate that at the original inside. That's f of x this time. And then I would need to multiply by the derivative of the inside function. Well the derivative of the inside function is f prime of x. Again I don't have a lot of information about f so when asked to say what its derivative is this is as far as I can go. If I wanted to clean this up slightly I could turn the negative exponent into a fraction and write this as negative f prime of x divided by f of x raised to the second power. So in other words just to kind of make a long story short here all the derivative stuff that we're doing here is pretty much the same as we have been doing. The only catch here is that we can't go as far as we used to go because we have these implicitly defined functions. When we encounter them in the derivative taking process we don't panic we just write f prime of x whenever we have to use the derivative. But we do take care to use correct rules here. The chain rule in this situation. The product rule in the one at the bottom and just a regular sum rule and power rule options for the first example. Thanks for watching.