If you haven't already given this problem a try on your own, go ahead and do that before checking back in with me. Alright. In this problem, we're asked to use implicit differentiation to find \( \frac{dy}{dx} \). And we're given here that the natural logarithm of \( x^2 \times y \) is equal to \( 2 \times e \) to the power of \( 3x + y \). So I'm going to start here by just rewriting my equation.
That's the natural logarithm of \( x^2 \times y \), and this is equal to \( 2e^{3x+y} \). So in order to use implicit differentiation to ultimately find \( \frac{dy}{dx} \), we need to take the derivative with respect to \( x \) of both sides here. So I'm going to start on this left-hand side. I want to take the derivative of the natural logarithm of \( x^2y \) with respect to \( x \). Now in doing this, this is going to give me \( \frac{1}{x^2y} \).
And then applying the chain rule, I then have to multiply this by the derivative of \( x^2y \). Now in order to find this derivative, we need to use the product rule because I have \( x^2 \) multiplying \( y \). So that means here that we need to apply our product rule. Remember our memory tool for the product rule, left \( d \) right plus a right \( d \) left. So I'm going to take that left-hand function \( x^2 \), multiply it by the derivative of that right-hand function \( y \), which is \( \frac{dy}{dx} \).
And I'm adding this together with that right-hand function \( y \) times the derivative of that left-hand function \( x^2 \), which is just \( 2x \). So this is my left-hand side, that whole derivative. This is then equal to the derivative of \( 2 \times e \) to the power of \( 3x+y \). That \( 2 \) is just a constant, so I can leave that out front. Then the derivative of \( e^{3x+y} \) is then \( e^{3x+y} \times \) the derivative of \( 3x+y \).
The derivative of \( 3x+y \), because we're using the chain rule here, is just going to be \( 3+\frac{dy}{dx} \). So now we have fully taken these derivatives, and we really just have to do a bunch of algebra from this point on because we want to find \( \frac{dy}{dx} \). So to ultimately get \( \frac{dy}{dx} \), since it's in 2 different terms, we need to first start by expanding this even further so that we can then simplify it down as we go on. So the first thing I want to do is actually just distribute everything into my parentheses here. So starting on this left-hand side, this gives me \( \frac{x^2}{x^2y} \times \frac{dy}{dx} + \frac{2xy}{x^2y} \).
Then this is equal to \( 6e^{3x+y} + 2e^{3x+y} \times \frac{dy}{dx} \). So now I can see that I have 2 terms that have this \( \frac{dy}{dx} \). I want to get those on the same side so that I can ultimately solve for that \( \frac{dy}{dx} \). So I want to pull this term over to this left side subtracting by it, and that means that I also want to take my terms that don't have \( \frac{dy}{dx} \) and put them on the same side as well.
So I'm going to subtract this term over to the right side. So cleaning this up a little bit as I go, I'm going to go ahead and cancel these \( x^2 \). So this gives me \( \frac{1}{y} \times \frac{dy}{dx} \). And then subtracting this term over here, this is \( -2e^{3x+y} \times \frac{dy}{dx} \). This is then equal to that \( 6e^{3x+y} - \frac{2}{x} \).
And I can do some cancellations here as well. Those \( y's \) cancel, and then one of my \( x's \) cancels as well. So this is just leaving me with \( \frac{2}{x} \) for that term. So looking at this left-hand side, I can factor out \( \frac{dy}{dx} \) since both of those terms have it. So that's \( \frac{dy}{dx} \times \left(\frac{1}{y} - 2e^{3x+y} \)
So all I've done is I've pulled that \( \frac{dy}{dx} \) out of both of those terms. Then on my right-hand side, there's nothing else that I need to do here yet. This is \( \frac{6e^{3x+y} - \frac{2}{x}} \). So now that I have this \( \frac{dy}{dx} \) factored out, in order to isolate it, all I need to do is get rid of this term, which I can do by dividing it over. So if I divide by it on both sides, so this is \( \frac{1}{y} - 2e^{3x+y} \).
That's going to cancel over here. So I'm dividing by it over here. This is \( \frac{1}{y} - 2e^{3x+y} \) again there. Then on my left-hand side, all that I'm left with finally is \( \frac{dy}{dx} \). But we definitely have a bit of algebra to do over here before this is fully simplified.
And in order to get through this algebra, since I have a fraction on top and I have this fraction on a bottom, but I want to go ahead and have my other term have a common denominator so that I actually can subtract them, and it will ultimately make it easier to simplify this further. So for this \( 6 \times e^{3x+y} \), I'm going to multiply this by \( \frac{x}{x} \) to give it that common denominator of \( x \). Then on the bottom, since this fraction, one over \( y \), has a denominator of \( y \), I'm going to multiply that other term by \( \frac{y}{y} \) to give it a common denominator as well. Now when we do this, this becomes \( \frac{6xe^{3x+y}-\frac{2}{x}} \) because I've given it that common denominator. And that's the numerator of this entire fraction.
And then on the denominator there, this then becomes \( \frac{1 - 2y \times e^{3x+y}}{y} \) because, again, I've given those a common denominator as well. Now here, since I'm dividing 1 fraction by another fraction, I am ultimately just multiplying this fraction by the reciprocal of this fraction. So what is this actually equal to? Well, we can rewrite this as \( \frac{6x \times e^{3x+y} - \frac{2}{x}} \). So that is my full numerator there.
And since I'm dividing by a fraction, I'm multiplying by the reciprocal of it. So this is then multiplied by \( \frac{y}{1 - 2y \times e^{3x+y}} \). So now I can just distribute this \( y \) into these two terms and distribute this \( x \) into these two terms to get my final answer. So here I have \( \frac{dy}{dx} \) is then equal to, distributing that \( y \) into those top two terms, \( \frac{6xy \times e^{3x+y} - \frac{2y}{x}} \) minus \( 2xy \times e^{3x+y} \). And this is my final simplified answer here.
Now I know that a lot of this algebra can be a little bit tedious to get through. Make sure that you're paying attention to every single term that is in each line as you work through these problems. I hope that this was helpful. And if you have any questions, let us know.
