Second Derivatives

In Exerci...

Question

Second Derivatives

In Exercises 19–26, use implicit differentiation to find dy/dx and then d²y/dx². Write the solutions in terms of x and y only.

2√y = x – y

Accepted Answer

Hello there. Today we're going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Find D to the power of 2 Y by DX to the power of 2 using implicit differentiation. 13 multiplied by the square root of Y is equal to X2 minus 3 Y. Awesome. So it appears for this particular problem, we're trying to find D squared Y per by, so it's gonna be D2Y per DX to the power of 2. So it appears we're trying to find the second derivative. For DY by DX using implicit differentiation, and we're applying it to this particular equation that's given to us by the prom itself. That's what that 2 means in our expression for D squared by D X squared. And we're trying to find the second derivative for DY by DX. So, with that said, Let us note first off once again that we are told that we're given 13 multiplied by the square root of Y is equal to X2 minus 3 Y. So our first step is that we need to figure out the first derivative, so we need to differentiate implicitly with respect to X. So we need to take 13. Divided by 2 multiplied by the square root of Y multiplied by Dy by DX is equal to 2 X minus 3 multiplied by Dy by DX. So now we need to solve for DY by DX. Of 13 divided by 2 multiplied by the square root of Y plus 3 is equal to 2 X, which will mean getting D Y by DX all by itself. You'll find that D Y by DX is going to be equal to drum roll, or X multiplied by the square root of Y divided by 13 + 6 multiplied by the square root of Y. So now we need to solve for the second derivative. So we need to differentiate Dy by DX using the quotient rule, so we need to take D to the power of 2 Y by DX to the power of 2. is equal to 13 + 6 multiplied by the square root of Y multiplied by parentheses 4 multiplied by the square root of Y plus 2 X divided by the square root of Y multiplied by Dy by D X C. Minus 4 X multiplied by the square root of Y, multiplied by 3 divided by the square root of Y, multiplied by DY by DX. All divided by parentheses 13 plus 6 multiplied by the square root of Y, all to the power of 2. Fantastic. So now our next step is we need to substitute D Y by D X is equal to 4 X multiplied by the square root of Y divided by 13 + 6 multiplied by the square root of Y. So we're going to be substituting this value in and simplifying. So when we do that, we will find that D to the power of 2 Y by DX to the power of 2 is going to be equal to parentheses 13 + 6 multiplied by the square root of Y, multiplied by parentheses 4 multiplied by the square root of Y plus 2 X divided by the square root of Y. Multiplied by 4 X multiplied by the square root of Y, divided by 13 + 6 multiplied by the square root of Y minus 4 X multiplied by the square root of Y multiplied by 3 divided by the square root of Y multiplied by. 4 X multiplied by the square root of Y, divided by 13 + 6 multiplied by the square root of Y. All divided by. Parentheses 13 + 6 multiplied by the square root of Y, all to the power of 2. Whew, that was a lot, but we're doing good. So now, what we need to do is we need to simplify our numerator first. So let us note that our first term is parentheses 13 + 6 multiplied by the square root of Y, multiplied by 4, multiplied by the square root of Y plus. 8 X to the power of 2. Divided by 13 + 6 multiplied by the square root of Y. Fantastic. So now we need to distribute parentheses 13 + 6 multiplied by the square root of Y, and when we do that, we will find that it's simply equal to parentheses 13 + 6 multiplied by the square root of Y, multiplied by 4 multiplied by the square root of Y plus 13 + 6 multiplied by the square root of Y. Multiplied by 8 X to the power to divided by 13 + 6 multiplied by the square root of Y. So now, we need to simplify each part, so we need to simplify parentheses 13 + 6 multiplied by the square root of Y, so 6. Multiplied by the square root of Y in parentheses, multiplied by 4 multiplied by the square root of Y, which is going to be equal to 52, and we're gonna take 52 multiplied by the square root of Y plus 24 Y. And we also need to simplify 13 + 6 multiplied by the square root of Y. Multiplied by 8 X to the power of 2 divided by 13 + 6 multiplied by the square root of Y, which is going to be equal to. Simply 8 X squared. Fantastic. So, moving right along here. So now we can note that the first term will simplified to 52 multiplied by the square root of Y plus 24 Y plus 8 X2. So our second term, we need to note, and let's make a note of this. So this is our second term, make a note of it in red. So, our second term is gonna be negative, 4 X multiplied by the square root of Y, multiplied by 3 divided by the square root of Y, multiplied by 4 X multiplied by the square root of Y, divided by 13 + 6 multiplied by the square root of Y. We now need to simplify our coefficients and exponents, so we need to take the square root of Y multiplied by 1 divided by the square root of Y, which will be equal to 1. So our expression will become negative 4x multiplied by 3, multiplied by 4 X multiplied by the square root of Y. Divided by 13 + 6 multiplied by the square root of Y, which will be all equal to -12 X multiplied by, so we're gonna be taking -12 X multiplied by. 4 X, so -12 X multiplied by 4 X multiplied by the square root of Y, and then we need to take 4 X multiplied by the square root of Y and we divide it by 13 + 6 multiplied by the square root of Y. Fantastic. So now we need to multiply our coefficients. So when we multiply our coefficients, we will get -48. So we're going to bring out the negative sign out front, so we'll get 48 X squared multiplied by the square root of Y, divided by 13 + 6 multiplied by the square root of Y. So now we need to combine the terms. So when we combine the terms, our numerator will become 52 multiplied by the square root of Y plus 24 Y plus 8 X2 minus 48. X squared multiplied by the square root of Y divided by 13 + 6 multiplied by the square root of Y, fantastic. So that means we now need to combine these terms and find a common denominator, which will be our common denominator will be 13 + 6 multiplied by the square root of Y. So that will mean, when we do that, we will find that it's equal to parentheses 52 multiplied by the square root of Y plus 24 Y plus 8 X2, so plus 8 X2 multiplied by 13 plus 6 multiplied by the square root of Y minus 48 X 2 multiplied by the square root of Y. And then this will be all divided by 13 + 6 multiplied by the square root of Y. Awesome. So in an effort to save some time, I'm gonna skip a like a lot of simplification process, but you're gonna use algebra to simplify this expression down to its most simple form, so you're just going to keep manipulating it until you can't simplify it anymore. And when you do that, the correct final answer for the most simplified form will be 676 multiplied by the square root of Y plus 624 Y plus 104 X squad, plus 144 Y to the power of three halves. All divided by 13 + 6 multiplied by the square root of Y, all to the power of 3. And that's it. That is our final answer. Hooray, we did it. We found our value for our final answer. So to quickly recap, looking at our problem itself, to quickly go over everything we went over together to quickly wrap everything up, pull everything together. Our final answer without a doubt has to be once again. 676 multiplied by the square root of y plus 624 Y plus 104 X to the power 2 plus 144 y to the power of three halves or 3 divided by 2, all divided by parentheses 13 + 6 multiplied by the square of Y to the power of 3. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Second Derivatives

In Exercises 19–26, use implicit differentiation to find dy/dx and then d²y/dx². Write the solutions in terms of x and y only.

2√y = x – y

Key Concepts

Implicit Differentiation

First Derivative (dy/dx)

Second Derivative (d²y/dx²)

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