Implicit differentiation problem.

In summary: Then factor out the ##\frac{dy}{dx}## and solve for it.In summary, in order to find dy/dx in terms of x and y, we start by deriving the left hand side of the given equation, x^2-sqrt(xy)+y^2=6. Then, we simplify the last term by multiplying it out and gathering all the terms with dy/dx onto one side and all other terms onto the other side. After factoring out dy/dx, we can solve for it to get the final answer.
  • #1
charmedbeauty
271
0

Homework Statement



Find dy/dx in terms of x and y if..

x2-√(xy)+y2=6





Homework Equations





The Attempt at a Solution



so I started by..


x2-√(xy)+y2=6

deriving the LHS

2x+2y(dy/dx)-1/2(xy)-1/2(1(y)+x(dy/dx))

Simplifying the last term

2x+2y(dy/dx)-(y+x(dy/dx))/(2√(xy))=6

taking the 2x over to separate dy/dx's

2y(dy/dx)-(y+x(dy/dx))/(2√(xy))= 6-2x

then I thought I would multiply through to get a common denominator..

[2y(dy/dx)(2√(xy))-(y+x(dy/dx))] / [2√(xy) =6 -2x

so multiply through by denominator to simplify and collect like terms

[2y(dy/dx)(2√(xy))-(y+x(dy/dx))] = (6-2x)(2√(xy))

so taking that -y over

[2y(dy/dx)(2√(xy))-(x(dy/dx))]= (6-2x)(2√(xy))+y

taking out dy/dx as a common factor

dy/dx[4y√(xy) - x] = (6-2x)(2√(xy))+y

so dy/dx = [(6-2x)(2√(xy))+y] / [4y√(xy) - x]

Is this right, because I checked it on wolfram and it had a different answer so I guess not, can someone please shed some light on where I went wrong?

Thanks.
 
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  • #2
charmedbeauty said:

Homework Statement



Find dy/dx in terms of x and y if..

x2-√(xy)+y2=6





Homework Equations





The Attempt at a Solution



so I started by..


x2-√(xy)+y2=6

deriving the LHS

2x+2y(dy/dx)-1/2(xy)-1/2(1(y)+x(dy/dx))

Simplifying the last term

2x+2y(dy/dx)-(y+x(dy/dx))/(2√(xy))=6

taking the 2x over to separate dy/dx's

2y(dy/dx)-(y+x(dy/dx))/(2√(xy))= 6-2x

then I thought I would multiply through to get a common denominator..

[2y(dy/dx)(2√(xy))-(y+x(dy/dx))] / [2√(xy) =6 -2x

so multiply through by denominator to simplify and collect like terms

[2y(dy/dx)(2√(xy))-(y+x(dy/dx))] = (6-2x)(2√(xy))

so taking that -y over

[2y(dy/dx)(2√(xy))-(x(dy/dx))]= (6-2x)(2√(xy))+y

taking out dy/dx as a common factor

dy/dx[4y√(xy) - x] = (6-2x)(2√(xy))+y

so dy/dx = [(6-2x)(2√(xy))+y] / [4y√(xy) - x]

Is this right, because I checked it on wolfram and it had a different answer so I guess not, can someone please shed some light on where I went wrong?

Thanks.

After the step labelled "Simplifying the last term" why to you have 6 on the right side?
 
  • #3
charmedbeauty said:

Homework Statement



Find dy/dx in terms of x and y if..

x2-√(xy)+y2=6





Homework Equations





The Attempt at a Solution



so I started by..


x2-√(xy)+y2=6

deriving the LHS

2x+2y(dy/dx)-1/2(xy)-1/2(1(y)+x(dy/dx))

Simplifying the last term

2x+2y(dy/dx)-(y+x(dy/dx))/(2√(xy))=6

Left hand side is good, but I'm not so sure you want to simplify the last term. What is ##\frac{d}{dx} (6)##?

taking the 2x over to separate dy/dx's

2y(dy/dx)-(y+x(dy/dx))/(2√(xy))= 6-2x

Every term on the left hand side isn't multiplied by ##\frac{dy}{dx}##, you need to move the other term over as well before you can factor ##\frac{dy}{dx}## out.

Try fixing those errors and continue.
 
  • #4
scurty said:
Left hand side is good, but I'm not so sure you want to simplify the last term. What is ##\frac{d}{dx} (6)##?



Every term on the left hand side isn't multiplied by ##\frac{dy}{dx}##, you need to move the other term over as well before you can factor ##\frac{dy}{dx}## out.

Try fixing those errors and continue.



ok yeah the 6 will cancel.

and when I take the 2x over I also take the -y over

"" " " "" = y-2x

??
 
  • #5
I'm not sure what you mean by the 6 canceling. The derivative of a constant is 0.

This just becomes an algebra exercise from here. The y isn't a term by itself, it is being multiplied by a number.

deriving the LHS

2x+2y(dy/dx)-1/2(xy)-1/2(1(y)+x(dy/dx))

Simplifying the last term

2x+2y(dy/dx)-(y+x(dy/dx))/(2√(xy))=6

Don't simplify the last term, multiply the last term out and then gather all the terms with ##\frac{dy}{dx}## onto one side and all the other terms onto the other side.
 

1. What is implicit differentiation?

Implicit differentiation is a method used in calculus to find the derivative of a function that is not explicitly written in terms of a single variable. It is commonly used when the dependent variable cannot be solved for in terms of the independent variable.

2. Why is implicit differentiation used?

Implicit differentiation is used when the dependent variable is expressed in terms of the independent variable in a way that makes it difficult or impossible to solve for the dependent variable. It allows us to find the derivative of the function without having to solve for the dependent variable first.

3. How is implicit differentiation done?

To perform implicit differentiation, we use the chain rule and the product rule to differentiate each term in the equation with respect to the independent variable. The dependent variable is treated as a function of the independent variable, and any instances of the dependent variable are replaced with its derivative.

4. What is an example of an implicit differentiation problem?

An example of an implicit differentiation problem is finding the derivative of the equation x^2 + y^2 = 25. Using implicit differentiation, we would differentiate each term with respect to x and replace any instances of y with dy/dx. This would result in the equation 2x + 2y(dy/dx) = 0. We can then solve for dy/dx to find the derivative of the function.

5. What are the applications of implicit differentiation?

Implicit differentiation has many applications in physics, engineering, and other fields where relationships between variables are not easily expressed in terms of a single variable. It is also used in optimization problems and in finding tangent lines to curves.

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