How do you use implicit differentiation to find #(dy)/(dx)# given #3x^2+3=ln5xy^2#?

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1 Answer
Euan S.
Aug 4, 2016

#(dy)/(dx) = y(3x - 1/(2x))#

Explanation:

Implicit differentiation is no different to normal, you just have to make sure to apply the chain rule when differentiating the #y# terms, ie #y# will differentiate to #(dy)/(dx)#.

#d/(dx)(3x^2 + 3 = ln(5xy^2))#

When we differentiate the log I used a combination of chain and product rules, but you could also rewrite it as :

#ln(5xy^2) = ln(5) + ln(x) + ln(y^2)#

using rules of logs and then just use chain rule.

#6x = 1/(5y^2x)*(5y^2 + 10xy(dy)/(dx))#

#30x^2y^2 = 5y^2 + 10xy(dy)/(dx)#

#5y^2(6x^2 - 1) = 10xy(dy)/(dx)#

#y/2(6x - 1/x) = (dy)/(dx)#

#therefore (dy)/(dx) = y(3x - 1/(2x))#

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