 Let's look at the derivatives of logarithmic and exponential functions. So our two key functions are e to the x, and log of x, and our derivatives are the following. Derivative of log is 1 over x, again in prime notation, meaning exactly the same thing, and also for e to the x. This is the world's simplest derivative. The derivative of e to the x is e to the x, and again prime notation e to the x prime is the same as e to the x. And the thing to remember is chain rule, chain rule, chain rule. So let's take the derivative of log of 3x plus 5. And again, we'll start out by a little bit of analysis. So I have my value of x times 3, add 5, hit it with the log, and so this type of function is a log function. So the derivative begins with the derivative of log, and again I'll drop out everything except for the last function. Derivative of log 1 over times chain rule derivative of whatever was there. And we apply the kindergarten rule, put things back where you found them, and now I need to find the derivative of 3x plus 5, and that's just going to be there. And there's my derivative, which we can simplify a little bit, but we're not going to do too much with. So there's my great simplification. Now what if I have something that looks more horrendous? Derivative of log of x squared plus 2x minus 7, and what do we have here? Again, take x squared, take x times 2, squared plus 2x, subtract 7, figure that out, then hit the whole thing with the log. This is once again a log function, so I'll drop everything except for that last thing that I do. Derivative of log 1 over times derivative of what was inside, and put things back where you found them. Derivative of x squared plus 2x minus 7, that's just a polynomial, so that's not really a difficult derivative to find. And we don't need to, but a nice simplification is to write that as a single fraction. Again, a useful check to make sure you're applying the derivative rules correctly. Anytime you're doing the derivative of anything more complicated than a polynomial, there's always going to be an echo of the original function. So again, here's our x squared plus 2x minus 7 in the original function. Here's our echo in the final function. How about an exponential function, e to the 5x? And again, a little bit of analysis goes a long ways. Take x times 5, e raised to that power. The last thing we do in this function is e raised to the power. So this is an exponential function. Drop everything except for the last thing that we do. The derivative of e to the is world's easiest derivative. Same thing, times, don't forget the chain rule, derivative of whatever was in there. Apply the kindergarten rule, put things back where you found them, and we do have to find the derivative of 5x, which is just going to be 5. So our echo of the original function, well in this case it's a pretty complete echo, e to the 5x here, e to the 5x down here. And it's really not too much different if we have a more complicated function. Here's e to minus x squared over 2. So again, take x, square it, change the sign, divide by 2, e raised to that number. And that means, again, we have our last function is e raised to the, and so when we want to find the derivative, we're going to start by dropping everything. This is an e to the type function. The derivative world's easiest derivative, same thing, times the derivative of whatever was in the parentheses. So put things back where you found them. I need to find the derivative of minus x squared over 2, which is the derivative of minus a half x squared, which is the derivative of, which is minus one half times 2x. And I might do a little bit of simplification to make things look a little bit nicer. Well, how about the derivative of 1 plus 3x raised to power x? So we don't have a derivative rule for this, but let's see what type of thing it is. Let's see. So I'm going to take x times 3 at 1, and I'm going to take x. The last thing I do is something to the x, and that looks and sounds a lot like e to the x. And the only difference here is I don't have an e to the x. I have a something to the x. But that does suggest that we might want to do a little bit of algebra and see if we can rewrite this function. And so one useful result from algebra is that any positive number can be written as e raised to the power of log of the same number. So that means I could rewrite 1 plus 3x to the x as e raised to the power of log of 1 plus 3x to the x. So I'll do that. That's 1 plus 3x to the x is the same thing as e raised to the log of that same number. Now, I can apply my rules of logarithms. I have the log of something to a power. And if I have that, I can simplify that exponent comes out in front. Note I have not done any differentiation yet. This is all algebra. But again, a little bit of analysis goes a long way. x times 3 plus 1. Take log of that. Hold it x times whatever that was. Figure that out. Then e raised to that power. This is an e to the something type of function. So my derivative e to the something. And again, world's easiest derivative. Derivative of e to the something. Same thing times the derivative. Put things back where you found them. So the thing that had been in the exponent there, x log 1 plus 3x. Put it back. Put it back. I want the derivative of this expression. Again, a little bit of analysis goes a long way. x times 3 plus 1. Hit it with a log times x. And there's your value of the function. The last thing that you do is multiply. So this is the derivative of a product. So I need to use the product rule. So I'll write down the product rule. That's the first times derivative of the second plus the second times the derivative of the first. I'll write it down. And again, this tells me what I need to do without actually doing it. But the derivative of 1 plus 3x derivative of x, those are things that I can do pretty easily. That's going to give me this. And we have the possibility. We have the option of doing some algebraic simplification. This is a good enough form as it is. And if you're in a rush, then this is a reasonable form to leave it in. However, if you hope to use this for something, it pays to do a little bit of algebraic simplification. And probably the two easy things to do, remember this horrible expression here. Originated because we needed to rewrite this expression. So one thing we can do, we can take this expression and rewrite it in its original form. And then x times 3, that's 3x. x times 3x plus 1 times 1, 1 plus 3x. And, well, that's not too challenging either, 3x plus 1 plus 3x. I can rewrite that. 6x plus 1, look for the echo. Original function had 1 plus 3x to the x. Derivative had 1 plus 3x to the x. We're probably going in the right direction.