 So this is a corrected version of this problem, where we want to find the derivative of 1 plus 3x raised to the power x. Now the thing to realize is that we don't actually have a formula for the derivative of a variable expression raised to a variable expression. So what can we do here? Well, we can always apply a little bit of algebra. So the key algebraic step here is that any positive value a is going to be the same as e raised to the power of the log of a. So I can rewrite this expression as an exponential. So here's my original expression 1 plus 3x to the x, and that's really the same as e to the log of 1 plus 3x to the x. So there's my first step, which is just the algebraic transformation of an expression into an equivalent exponential expression. And why bother hitting things with a log if you're not going to take advantage of the fact that logs transform complicated powers and products and quotients into things that are a lot simpler. In particular, this is the log of something raised to the x power. So I'm going to rewrite this by bringing that exponent out to the front. And that gives me the derivative of an expression, which if I look at it, this is an e to the expression. So I'm going to drop everything but the last thing that I do, and we now have world's ECS derivative. The derivative of e to the who cares what is going to be the same thing times the derivative of whatever our exponent was. And we apply the kindergarten rule. We'll put everything back where we found it. Now to continue this derivative, we need to find the derivative of x times log of 1 plus 3x, and that's a product. So that's going to be first times derivative of the second plus the second times derivative of the first. And I can differentiate those here. Derivative of x is just 1. Derivative of log is 1 over times derivative of the inside. And so after all the dust settles, I get that as my derivative, log 1 plus 3x. And I'll do a little bit of algebraic simplification here. I can multiply it by 1 and not change anything. I can multiply 3 by x and get something a little bit simpler. But I'm not going to do a lot of algebraic simplification. Finally, one last check. Remember that if you're doing a derivative more complicated than x to the n, there will always be an echo of the original function. So here my original function, 1 plus 3x to the x, and here's my echo, 1 plus 3x also to the x.