Differentiation using multiple rules
Derivative of sin(ln(x²))
So now we're going to attempt to take the derivative of the sine of the natural log of x squared. So now we have a function that's the composite of a function, that's a composite of another function. So one way you could think of it, if you set f of x as being equal to sine of x, and g of x being the natural log of x, and h of x equaling x squared. Then this thing right over here is the exact same thing as trying to take the derivative with respect to x of f of g of h of x. And what I want to do is kind of think about how I would do it in my head, without having to write all the chain rule notation. So the way I would think about this, if I were doing this in my head, is the derivative of this outer function of f, with respect to the level of composition directly below it. So the derivative of sine of x is cosine of x. But instead of it being a cosine of x, it's going to be cosine of whatever was inside of here. So it's going to be cosine of natural log-- let me write that in that same color-- cosine of natural log of x squared. I'm going to do x that same yellow color. And so you could really view this, this part what I just read over here, as f prime, this is f prime of g of h of x. This is f prime of g of h of x. If you want to keep track of things. So I just took the derivative of the outer with respect to whatever was inside of it. And now I have to take the derivative of the inside with respect to x. But now we have another composite function. So we're going to multiply this times, we're going to do the chain rule again. We're going to take the derivative of ln with respect to x squared. So the derivative of ln of x is 1/x. But now we're going to have 1 over not x, but 1 over x squared. So to be clear, this part right over here is g prime of not x. If it was g prime of x, this would be 1 over x. But instead of an x, we have our h of x there. We have our x squared. So it's g prime of x squared. And then finally, we can take the derivative of our inner function. Let me write it. So we could write this is g prime of h of x. And finally, we just have to take the derivative of our innermost function with respect to x. So the derivative of 2x with respect to-- or the derivative of x squared with respect to x is 2x. So times h prime of x. Let me make everything clear. So what we have right over here in purple, this, this, and this are the same things. One expressed concretely, one expressed abstractly. This, this, and this are the same thing, expressed concretely and abstractly. And then finally, this and this are the same thing, expressed concretely and abstractly. But then we're done. All we have to do to be done is to just simplify this. So if we just change the order in which we're multiplying, we have 2x over x squared. So I can cancel some out. So this x over x, 2x over x squared is the same thing as 2 over x. And we're multiplying it times all of this business. So we're left with 2 over x. And this goes away. 2 over x times the cosine of the natural log of x squared. So it seemed like a very daunting derivative. But we just say, OK, what's the derivative of sine of something with respect to that something? Well, that's cosine of that something. And then we go in one layer, what's the derivative of that something? Well, in that something we have another composition. So the derivative of ln of x, or ln of something with respect to another something, well that's going to be 1 over the something. So we had gotten a 1 over x squared here, that squared got canceled out. And then finally, the derivative of this innermost function, it's kind of like peeling an onion. The derivative of this inner function with respect to x, which was just 2x. Which we got right over here. This was 1 over x squared. This was 2x before we did any canceling out. So hopefully that helps clear things up a little bit.