Find d y / d x dy/dx d y / d x for the equation below using implicit differentiation. 3 y = x − y 3\sqrt{y}=x-y 3 y <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z"> = x − y

we need to find the derivative of y with respect to x using the process of implicit differentiation. Now what I can see is that we have y that shows up twice in this equation. And since we have y within the equation here, we're going to need to use the chain rule eventually to find dy over dx, but let's just start by taking the derivative on both sides of this equation. So doing this, I'll first have the derivative with respect to x of 3 times the square root of y. Then that whole thing is going to be equal to the derivative with respect to x of x, and it's going to be minus as we have this minus sign here, the derivative with respect to x of y. So we have a lot of steps here, so let's start with this first piece. Now for the derivative with respect to 3 times the square root of y, I'm actually going to change this up. So we're gonna focus a little bit on this on the square root of y here by saying this is the same thing as finding the derivative with respect to x of 3 times y to the one half power. Because the one half power is the same as having the derivative of the square root of y, but this time, we can now see it's a power. So we clearly see where we can incorporate this power rule when we solve this problem. Now the first thing I'm going to do is take this power of 1 half and multiply it by the front, then I'm gonna reduce the power by 1. That's just using the power rule, except we're using it on this 1 half power here, but we can still do that. So what that means is we're going to have 3 times 1 half, which is 3 halves, and we're going to have y. And then to the one half minus 1, that's just gonna give you negative one half. Because like if you're subtracting a whole from a half, you'll get a negative half. So that's what you get here. Now this whole thing, we can also write since we have to the negative one half power, we can also write this as 3 over 2 times 1 over y to the one half power. You can always write these negative powers as bringing it down to the bottom and writing it as a positive power. So you're gonna take the reciprocal here, which is what we just did. And then remember to the one half power is the same as the square root. So the whole thing is gonna become 3 over 2 times the square root of y. I just combine these two fractions by multiplying the tops and bottoms, and I change this to a square root. So this whole thing becomes 3 halves or or 3 over 2 times the square root of y. But remember, we need to also use the chain rule here. So we're taking the derivative of y with respect to x. So we have to multiply this entire thing by dy overdx. So this right here would be the result that we get for this derivative up here. So this entire process led us to this result here. So that's going to be what we do when we take this first derivative, and that's probably the trickiest derivative that we have here. So I'll get rid of all of this and we're just gonna move this up here. So we have the our derivative of the 3 times the square root of y with respect to x is this expression that we just calculated. So that's gonna be our first piece here. Now, I still need to find the derivative of everything on the right side of this equal sign, but thankfully that's a little bit more straightforward. Because the derivative of x with respect to x is just gonna give you 1. And then the derivative of y with respect to x, well, the derivative of y would just be 1, but then multiplying it by the derivative of y with respect to x, well, that whole thing is just gonna give you d y over d x. Right? So this is gonna be our entire expression right here, and we're trying to find the dy over dx. Well, since this since this is what we're trying to find, I'm just gonna use some algebra to get this term by itself. I'm gonna add dy over dx to both sides of the equation. Now what I'm trying to do in this step is I'm trying to get both dy over d x's on the same side of the equal sign. So notice that this will cancel here because we have the positive and the negative that cancel each other, but I can see that we'll have 2 dy's over d x's on this left side here, which is good. So rewriting this expression, I'll get 3 over 2 times the square root of y, and that's gonna be plus, and then we have d y over d x plus d y over d x's. Right? This is just a term like we've seen in algebra. Now this whole thing is equal to 1. Now from here, what I'm going to do is I'm going to subtract this 3 over 2 times the square root of y on both sides of the equation. So I'm subtracting this entire thing, and that's going to get it to cancel on the left side. So if we do this right, all we should end up with is 2 dy over dx on the left side of the equation. So that's gonna be this term. These are being multiplied. And the whole thing is going to be equal to what we have on the right side here, which is 1 minus 3 over 2 times the square root of y. Now there's still one more step I need to do in spite of all of this work that we just did because there's still a 2 that's in front of this dy over dx. So you need to divide that 2 through on both sides of the equation. That's gonna be in essence just multiplying everything on this right side here by 1 over 2. That's the same as dividing this whole thing by 2. So what that's going to mean is that the twos will cancel right here, meaning all we're going to end up with is dy overdx, and that's the term that we are trying to isolate on one side. So it's good that we got this by itself. But now I need to set this whole thing equal to everything that we have on the right side here. And if I go ahead and distribute this one half into these brackets here, well, I'm gonna have one half of 1, which is just one half minus and then one half times this expression, which will just multiply that 2 by that 2 on bottom. That will be combining these fractions here, which will give us 3 over 4 times the square root of y. So this right here would be the answer for dy overdx. Now another thing you might be asked to do is simplify this further by combining these 2 into 1 fraction. I mean, that's that's not necessarily always gonna be the case, so this might be a sufficient answer depending on who your professor is and what class you're in. But if you are asked to simplify this fully, well, what we can do is try to combine these as one fraction. So I would take this left fraction here, the one half, and I would multiply the top and bottom by 2 times the square root of y. So what that'll do is gonna be a 2 square root y on top, and then 2 times 2 would be 4 square root y on bottom. And this just allows me to have the same denominator that I have there. So we'd have minus 3 over 4 times square root of y. And remember, you can multiply the top and bottom by 2 square root y. That's perfectly a reasonable step to do as long as you're doing it to the top and bottom of the fraction when you multiply here. So that was the step that we did there. And what that allows us to do is combine the top and bottom equation. So we would get 2 square root y minus 3 on top of this. Since we have everything the same on bottom, we would get 4 square root y on bottom. So this would be another way that you could express your answer here if you needed to have one fraction as your result. But again, both of these would be the correct answer to this problem and give you the solution for d y over d x. So that's how you can solve these types of problems. Let's go ahead and move on.

4. Applications of Derivatives

Implicit Differentiation

Calculus

4. Applications of Derivatives

Implicit Differentiation

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Multiple Choice

Find $dy/dx$ for the equation below using implicit differentiation.
$3$\sqrt{y}$=x-y$

$dy/dx=3$\sqrt{y}$-x$

$dy/dx=$\frac{2\sqrt{y}$-3}{4$\sqrt{y}$}$

$dy/dx=$\frac{2\sqrt{y}$}{3+2$\sqrt{y}$}$

$dy/dx=$\frac{3}{3\sqrt{y}$-2xy}$

Verified step by step guidance

Start by differentiating both sides of the equation 3√y = x - y with respect to x. Remember that y is a function of x, so you'll need to use implicit differentiation.

Differentiate the left side: The derivative of 3√y with respect to x involves the chain rule. Let u = √y, then 3u is differentiated as 3 * (1/2√y) * dy/dx.

Differentiate the right side: The derivative of x with respect to x is 1, and the derivative of y with respect to x is dy/dx.

Set the derivatives equal: Combine the derivatives from both sides to form the equation 3 * (1/2√y) * dy/dx = 1 - dy/dx.

Solve for dy/dx: Rearrange the equation to isolate dy/dx on one side. This involves combining like terms and factoring out dy/dx.

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Struggling with Calculus?

Find dy/dxdy/dxdy/dx for the equation below using implicit differentiation.3y=x−y3\(\sqrt{y}\)=x-y3y=x−y

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Implicit Differentiation practice set

Find $dy/dx$ for the equation below using implicit differentiation.
$3\(\sqrt{y}\)=x-y$