How do you implicitly differentiate #3y + y^4/x^2 = 2#?

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How do you implicitly differentiate #3y + y^4/x^2 = 2#?

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Answer 1 · 2016-07-07T17:08:18

I found: #(dy)/(dx)=(2y^4)/(3x^3+4xy^3)#

Explanation:

ou need to remember to derive also #y# as a function of #x#, so, for example:
#y^2# derived will give you: #2y(dy)/(dx)#
in your case you get:
#3(dy)/(dx)+(4y^3x^2(dy)/(dx)-2xy^4)/x^4=0#
#3(dy)/(dx)+4y^3/x^2(dy)/(dx)-2y^4/x^3=0#
collect #(dy)/(dx)# and rearrange:
#(dy)/(dx)=(2y^4/x^3)/(3+4y^3/x^2)=(2y^4)/(3x^3+4xy^3)#

Answer 2 · 2016-07-07T17:13:26

yes

Explanation:

me too when it was asked elsewhere

https://socratic.org/questions/how-do-you-implicitly-differentiate-3y-y-4-x-2-2-2#285710

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How do you implicitly differentiate #3y + y^4/x^2 = 2#?

Thanks for your help!

2 Answers

Explanation:

Explanation:

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