Implicit Differentiation Watch

debbie394
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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?
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the bear
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(Original post by 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?
y2 + 2xlny = x2

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

let W = y2

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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Lilith2400
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(Original post by 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?
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

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debbie394
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(Original post by 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

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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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Lilith2400
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(Original post by esmeralda123)
I understand now. Thank you
But how would I know when to differentiate implicitly? What do I have to look out for?
You will have an equation which will be a mixture of two variables such as y and x. In normal differentiation the y is normally alone on one side of the equation and all of the x's are on the other side but in equations where the x's and y's are mixed up it may be too difficult rearrange them to get an equation where y= x so using implicit differentiation is easier
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timif2
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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)
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debbie394
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(Original post by 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)
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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MR1999
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(Original post by 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
Use the properties of log to rewrite the function.

\ln \dfrac{a}{b}= \ln a - \ln b

 \ln a^b = b\ln a

Spoiler:
Show






You should get:

 y= \ln \sqrt{\dfrac{2x-1}{2x+1}} =\dfrac{1}{2}\ln (2x-1) - \dfrac{1}{2}\ln (2x+1)

and then differentiate this normally using the chain rule.




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