# Implicit Differentiation

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Someone pleaseeee help. I dread implicit differentiation!!

Question : y2 + 2x ln y = x2

A curve has IMPLICIT EQUATION y2 + 2x ln y = x2 .

Verify that the point (1, 1) lies on the curve, and find the gradient of the curve at this point.

Answer:

2 y dx 2 ln y 2 x. y . dx 2 x

My Questions:

1)Why implicity differentiate when it says in the question above

2)When differentiating why do you have to differentiate 2x ln y separately so that there are two terms for them

3)How do you know when to put the dy/dx next to the terms. Is it only when you differentiate a term with a y?

4)Is it dy/dx so respect with x because that is at the end of the equation after the =. What if this is not clear, how do I know what goes on the bottom of the fraction dy/dx?

Question : y2 + 2x ln y = x2

A curve has IMPLICIT EQUATION y2 + 2x ln y = x2 .

Verify that the point (1, 1) lies on the curve, and find the gradient of the curve at this point.

Answer:

2 y dx 2 ln y 2 x. y . dx 2 x

My Questions:

1)Why implicity differentiate when it says in the question above

2)When differentiating why do you have to differentiate 2x ln y separately so that there are two terms for them

3)How do you know when to put the dy/dx next to the terms. Is it only when you differentiate a term with a y?

4)Is it dy/dx so respect with x because that is at the end of the equation after the =. What if this is not clear, how do I know what goes on the bottom of the fraction dy/dx?

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Someone pleaseeee help. I dread implicit differentiation!!

Question : y2 + 2x ln y = x2

A curve has IMPLICIT EQUATION y2 + 2x ln y = x2 .

Verify that the point (1, 1) lies on the curve, and find the gradient of the curve at this point.

Answer:

2 y dx 2 ln y 2 x. y . dx 2 x

My Questions:

1)Why implicity differentiate when it says in the question above

2)When differentiating why do you have to differentiate 2x ln y separately so that there are two terms for them

3)How do you know when to put the dy/dx next to the terms. Is it only when you differentiate a term with a y?

4)Is it dy/dx so respect with x because that is at the end of the equation after the =. What if this is not clear, how do I know what goes on the bottom of the fraction dy/dx?

**esmeralda123**)Someone pleaseeee help. I dread implicit differentiation!!

Question : y2 + 2x ln y = x2

A curve has IMPLICIT EQUATION y2 + 2x ln y = x2 .

Verify that the point (1, 1) lies on the curve, and find the gradient of the curve at this point.

Answer:

2 y dx 2 ln y 2 x. y . dx 2 x

My Questions:

1)Why implicity differentiate when it says in the question above

2)When differentiating why do you have to differentiate 2x ln y separately so that there are two terms for them

3)How do you know when to put the dy/dx next to the terms. Is it only when you differentiate a term with a y?

4)Is it dy/dx so respect with x because that is at the end of the equation after the =. What if this is not clear, how do I know what goes on the bottom of the fraction dy/dx?

^{2}+ 2xlny = x

^{2}

since there is a mixture of y and x you have to use implicit differentiating...

let W = y

^{2}

you want to find dW/dx....

dW/dx = dW/dy*dy/dx

dW/dy = 2y

so dW/dx = 2y*dy/dx

let M = 2x*lny

to find dM/dx we use the product rule

M = u*v where u = 2x and v = lny

du/dx = 2

dv/dx = dv/dy*dy/dx = {1/y}*dy/dx

so putting these bits together for the product rule:

dM/dx = lny*2 + 2x*{1/y}*dy/dx

the RHS is easier... you get just 2x when you differentiate.

====> 2y*dy/dx + 2lny + 2x*{1/y}*dy/dx = 2x

make dy/dx the subject.

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**esmeralda123**)

Someone pleaseeee help. I dread implicit differentiation!!

Question : y2 + 2x ln y = x2

A curve has IMPLICIT EQUATION y2 + 2x ln y = x2 .

Verify that the point (1, 1) lies on the curve, and find the gradient of the curve at this point.

Answer:

2 y dx 2 ln y 2 x. y . dx 2 x

My Questions:

1)Why implicity differentiate when it says in the question above

2)When differentiating why do you have to differentiate 2x ln y separately so that there are two terms for them

3)How do you know when to put the dy/dx next to the terms. Is it only when you differentiate a term with a y?

4)Is it dy/dx so respect with x because that is at the end of the equation after the =. What if this is not clear, how do I know what goes on the bottom of the fraction dy/dx?

2.) 2xlny is two functions multiplied together so to differentiate this you have to use the product rule. If 2xlny = uv , d(uv)/dx = udv/dx + vdu/dx which splits it up into two terms as you can see by the formula above.

3.) Everytime you differentiate a y term you must times it by dy/dx afterwards.

4.) I am not sure what you mean by this question sorry

Posted from TSR Mobile

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(Original post by

1.) I'm not exactly sure that I know what you mean in this question but you differentiate the equation because it wants you to find the gradient and when you differentiate and sub in a point it will give you the gradient of the curve at this point.

2.) 2xlny is two functions multiplied together so to differentiate this you have to use the product rule. If 2xlny = uv , d(uv)/dx = udv/dx + vdu/dx which splits it up into two terms as you can see by the formula above.

3.) Everytime you differentiate a y term you must times it by dy/dx afterwards.

4.) I am not sure what you mean by this question sorry

Posted from TSR Mobile

**Lilith2400**)1.) I'm not exactly sure that I know what you mean in this question but you differentiate the equation because it wants you to find the gradient and when you differentiate and sub in a point it will give you the gradient of the curve at this point.

2.) 2xlny is two functions multiplied together so to differentiate this you have to use the product rule. If 2xlny = uv , d(uv)/dx = udv/dx + vdu/dx which splits it up into two terms as you can see by the formula above.

3.) Everytime you differentiate a y term you must times it by dy/dx afterwards.

4.) I am not sure what you mean by this question sorry

Posted from TSR Mobile

But how would I know when to differentiate implicitly? What do I have to look out for?

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(Original post by

I understand now. Thank you

But how would I know when to differentiate implicitly? What do I have to look out for?

**esmeralda123**)I understand now. Thank you

But how would I know when to differentiate implicitly? What do I have to look out for?

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#6

It’s actually quite simple, when you have two variables x and y unknown, you have to use implicit

Anytime you differentiate any y variable, just pop a dy/dx next to it

Product rule is just u’(v) + v’(u)

Anytime you differentiate any y variable, just pop a dy/dx next to it

Product rule is just u’(v) + v’(u)

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(Original post by

It’s actually quite simple, when you have two variables x and y unknown, you have to use implicit

Anytime you differentiate any y variable, just pop a dy/dx next to it

Product rule is just u’(v) + v’(u)

**timif2**)It’s actually quite simple, when you have two variables x and y unknown, you have to use implicit

Anytime you differentiate any y variable, just pop a dy/dx next to it

Product rule is just u’(v) + v’(u)

Question 3 of http://pmt.physicsandmathstutor.com/...t%201%20QP.pdf

Answer: http://pmt.physicsandmathstutor.com/...t%201%20MS.pdf

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#8

(Original post by

Ok, i am doing a set of questions on implicit differentiation and i have come across this question, but it does not seem that implicit differentiation has been used because y is known. What type of differentiation has been used and how do i answer it. I have looked at the answer but don't understand it.

Question 3 of http://pmt.physicsandmathstutor.com/...t%201%20QP.pdf

Answer: http://pmt.physicsandmathstutor.com/...t%201%20MS.pdf

**esmeralda123**)Ok, i am doing a set of questions on implicit differentiation and i have come across this question, but it does not seem that implicit differentiation has been used because y is known. What type of differentiation has been used and how do i answer it. I have looked at the answer but don't understand it.

Question 3 of http://pmt.physicsandmathstutor.com/...t%201%20QP.pdf

Answer: http://pmt.physicsandmathstutor.com/...t%201%20MS.pdf

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