 Hi everyone, I have for you here eight examples of how to use the chain rule for finding derivatives. So maybe after the first few problems you could pause this video as you watch it to try the problem on your own before you start it again to see how it works out. So remember the whole idea with the chain rule is you essentially start on the outside and work your way in taking derivatives as you go. So in the first one the outermost function is the part that's raising that 7x plus 8 to the fourth power. So that's where we start. That's almost like applying the power rule. So the four goes to the front, we keep the 7x plus 8. That part's really important, that's actually the part a lot of people do wrong is they take too many derivatives at once, one derivative at a time. Remember that, one derivative at a time. So that's the first part of the chain rule. Now we're going to go inside and multiply by the derivative of the 7x plus 8. So that's just going to be 7. And then we just simplify as we're able. So here we can multiply the 4 and the 7 together to make 28. And we have 28 times the quantity 7x plus 8 cubed and that would be our answer. Now for this next one we're going to want to think of the square root function as being raised to the one half power. And then it's very similar to the first problem we looked at. So once again we start on the outside and work our way in. So the one half goes to the front and remember it's one derivative at a time so we're going to keep this argument here for now. And now it's going to be raised to the negative one half because remember you need to take one away from that exponent. Now we're going to go inside and multiply by the derivative of this. And we do need that in parentheses because it is the entire quantity that we want to multiply by. So we can really just leave our answer as that. If you want to rewrite it you're welcome to. You could have 6x minus 1. We could take that square root quantity to the denominator if you'd like. But otherwise simply leaving it as we had it is perfectly acceptable. So let's look at some more. So this is another one we will want to rewrite. Since we have a radical of a quantity raised to a power we can think of this as 3x squared plus 2 raised to the three fifths. And then once again it's very very similar to the earlier problems. So starting on the outside three fifths comes to the front. To keep the 3x squared plus 2 we need to subtract 1 from the exponent. So that gives us an exponent now of negative two fifths. So now we want to go inside and we're going to multiply by the derivative of the 3x squared plus 2. So that would be 6x. Again we can simplify it just to make it look nice. So we could say it's 18 over 5x. And we can leave that quantity as raised to a negative exponent that's perfectly fine. If you feel better bringing it to the denominator so as to rewrite it with a positive exponent that's fine as well. So as long as you remember a few guidelines the chain rule really isn't that hard. You want to remember one derivative at a time and just start on the outside and work your way in. Now this is another one that you looking at it as it is you might think to use the quotient rule which you very much could do here in this case. But let's talk about how you could rewrite it if we wanted to turn it into a chain rule problem. What we could do is bring that denominator quantity up to the numerator. And in doing so we now have it raised to the negative third exponent. So now it becomes essentially a classic chain rule problem. So we have the negative three exponent times the negative four in the front. That makes 12. Remember keep the 4x minus one, one derivative at a time. Take away one from the exponent and now we have negative four. Now we go inside and multiply by the derivative of the 4x minus one, which would be 4. We can multiply together the 12 and the 4 to make 48. And once again it's perfectly acceptable to leave this answer as having a negative exponent. However if you wish to bring that quantity to the denominator making it a positive four exponent that's fine as well. Now we have a couple more trig functions. So let's talk about this first one. We have sine squared of x. Now getting started now and being new to using the chain rule. You want to remember what sine squared of x means. It really means that we have sine of x that quantity squared. You might find it helpful to rewrite it like this the first few times while you're still getting used to using the chain rule. Now that it's rewritten like this it again becomes a classic chain rule problem. So taking one derivative at a time we're going to start with the outside squared function that comes to the front. Keep the sine of x and now it's essentially raised to the first power. If you find it helpful to put it there you can otherwise you can just kind of omit it that's fine. And now we need to multiply by the derivative of the inside function, derivative of sine, which of course is cosine. Now looking at that hopefully it hits you in the head. This is a double angle identity, which is sine of 2x. Again, double angle identities and knowing those definitely comes in handy. So let's look at number six, another trig one. This is a great example of a composite function. The inside function being the negative 2x to the fifth. The outside function being the cosine. Alright so we're going to start with by taking the derivative of the cosine part which we know is to be negative sine. Now remember though we take only one derivative at a time. So we're going to leave it as negative 2x to the fifth. Now we need to multiply that by the derivative of the negative 2x to the fifth. Which is negative 10x to the fourth. Alright so we can make it a little bit better. We have a negative times a negative. So that makes positive. We can throw this 10x to the front and we would have sine of negative 2x to the fifth. And that would be our answer. Let's look at a couple of other examples of trig equations. So once again the outside function is cosine and this one has three functions. We have a cosine, we have a sine and a 4x. So many embedded functions here but still we're going to start on the outside. Derivative of cosine is negative sine. And for now we keep the sine of 4x. Remember our rule, one derivative at a time. And now we're going to take the derivative of the sine part. Alright and we know derivative of sine is cosine. And we just keep the 4x, the argument 4x. Now we have one more to do. And now we have to do the derivative of the 4x, which is just 4. Alright so we can multiply it to just clean it up a little bit. We can put the 4 coefficient in the very front with the negative. It doesn't really matter whether the sine goes first or the cosine. Some people like to do cosine first, doesn't really matter. And this is how our final answer would look. Again you want to remember our guidelines. One derivative at a time working from the outside in. So our last example, yet another trig function. And this is another one we're going to rewrite because it's secant to the fourth. So think of this as secant of 2x squared. That entire quantity to the fourth power. Alright so really we have three embedded functions. The very outermost function being raised to the fourth power. And then we have the secant and we have the 2x squared. So we're going to start on the outside and work our way in. So the four comes to the front. We're going to keep the argument secant 2x squared. And we have to take one away from that exponent. So that's just the first part of the chain rule. Now we're going to go inside and do derivative of secant. Now remember derivative of secant is secant times tangent. We're going to keep the 2x squared argument as we do that. So we have secant 2x squared tangent 2x squared. Now finally we need the derivative of the 2x squared part which is simply 4x. Now if you like putting these in parentheses to offset them please do feel free. A lot of people find that helpful. So let's see what we can do to clean this up a little bit. We have a 4 times 4 to make 16. We can put that x in the very front with it. Notice how many secants we have. We seem to have four of them because we have that secant cubed but then we have times another secant. So what we really have is secant to the fourth of 2x squared. And then we have that tangent of 2x squared. Now if you prefer to write this part, if it just seems to help you out a little bit more. As like the quantity to the fourth power that's perfectly acceptable as well. So hopefully this gives you some really good idea of how to work with the chain rule in order to find derivatives of functions.