I have the x to the third plus x squared. And this thing right over here happens to be the derivative of x to the third plus x squared. The derivative of x to the third is three x squared. Derivative of x squared is two x, which is a huge clue to me that I could use u-substitution. So what I do here is this thing, or this little expression here, where I also see its derivative being multiplied, I can set that equal to u. So I can say u is equal to x to the third plus x squared. Now what is going to be the derivative of u with respect to x?