In this problem, we're asked to find the derivative of the given function, and the function that we're given here is \( y = 2x^3 - x^2 + 4x + 7 \). Now, I'm going to fully break down this derivative with all of the rules for derivatives that we know. Now as you get better and better at taking derivatives, you're not going to have to break it down fully in this way. But for now, I'm going to show you the entire process of taking this derivative. So let's go ahead and get started.
Now here we have our function \( y \). So our derivative here, we could denote as \( \frac{dy}{dx} \) because we want to take the derivative with respect to this variable \( x \). Now, you could also denote this derivative as \( y' \), and that would still be accurate. Now taking our derivative here, \( \frac{dy}{dx} \), I have a bunch of different terms being added together, and I also have some multiplication of some constants happening. So in order to fully break this down using all of the rules for derivatives that we know, we're going to go ahead and separate all of these out.
Now we know that whenever we have a constant, we can already pull that out to the front of our derivative. So when thinking about this first term here, \( 2x^3 \), I can think of this derivative as \( 2 \times \frac{d}{dx} of x^3 \), and I'll eventually take that derivative using the power rule. Now for my next term, I have a minus \( x^2 \). So here, using my sum and difference rule, specifically here, a difference, I am subtracting here the derivative \( \frac{d}{dx} \) of that \( x^2 \). Then for my next term, I have plus \( 4x \).
Since I have another constant, I can think of this as pulling that \( 4 \) out to the front of my derivative, and this is plus \( 4 \times \frac{d}{dx} \) of \( x \). Then lastly, I have a constant down here. So I have plus \( \frac{d}{dx} \) of \( 7 \). Now I can think of all of these derivatives individually and add and subtract them as needed. Now here, looking at this first derivative here, \( 2 \times \frac{d}{dx} \) of \( x^3 \).
I already have that \( 2 \). I want to keep that as is, that constant. And then using the power rule here, I want to go ahead and pull that \( 3 \) out to the front and then decrease that power by 1. That's using the power rule that we just learned. So this becomes \( 3 \times x^{3-1} \), which is just \( 2 \).
Now from here, we can take our next derivative. Here, I'm subtracting this derivative, remember? And again, I'm going to be using the power rule here. Now with this \( x^2 \) I'm going to pull that \( 2 \) out to the front and then decrease that exponent, that power, by 1. So this becomes \( 2 \times x^{2-1} \), which is just \( 1 \) so we don't have to worry about writing it.
Then adding my next derivative, I have plus \( 4 \times \frac{d}{dx} \) of \( x \). Now we know that the derivative of \( x \) is just \( 1 \), so this is really just \( 4 \times 1 \). Then lastly, I have the derivative of my constant down here, \( \frac{d}{dx} \) of \( 7 \). I'm adding that together here. Now the derivative of any constant by itself is just \( 0 \), so I'm just adding a \( 0 \) on there.
Now we can do some more simplification here by just multiplying some stuff out and taking away that \( 0 \) on the end. So multiplying this term out, \( 2 \times 3x^2 \), it gives me \( 6x^2 \), and then I have a minus \( 2x \) that next term, plus \( 4 \times 1 \), which is just \( 4 \), And this gives me my final derivative \( \frac{dy}{dx} \) of that original function. Now remember, like I said, as you continue taking more and more derivatives, you're really going to be able to skip right from your original function to your derivative without having to write out all of those steps. Now as we get there, we're going to continue to get more practice. So feel free to let me know if you have any questions, and I'll see you in the next one.
