Using implicit differentiation, find (dy)/(dx) x^3y^6=(x+y)^9

Krish Schmitt 2022-10-08 Answered
Using implicit differentiation, find d y d x
x 3 y 6 = ( x + y ) 9
My Attempt:
x 3 y 6 = ( x + y ) 9
Differentiating both sides with respect to x, we get:
d d x ( x 3 y 6 ) = d d x ( ( x + y ) 9 )
x 3 . d d x y 6 + y 6 . d d x x 3 = 9 ( x + y ) 8 d d x ( x + y )
6 x 3 y 5 . d y d x + 3 x 2 y 6 = 9 ( x + y ) 8 ( 1 + d y d x )
d y d x = 9 ( x + y ) 8 3 x 2 y 6 x 3 y 5 9 ( x + y ) 8
But the answer is
d y d x = y x
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Answers (2)

Houston Ellis
Answered 2022-10-09 Author has 4 answers
I fixed a typo in your question, and with that, your implicit differentiation is perfectly correct. However, it looks difficult (as in, I cannot see how) to blackuce it from that to the expected simple answer.

However, if you're dealing only with the real domain it makes things a lot easier to take cube roots right at the outset. This simplifies things greatly.
x 3 y 6 = ( x + y ) 9
For x , y R ,
x y 2 = ( x + y ) 3
Differentiating implictly wrt x,
y 2 + 2 y y x = 3 ( x + y ) 2 ( 1 + y )
y ( 2 y x 3 ( x + y ) 2 ) = 3 ( x + y ) 2 y 2
y = 3 ( x + y ) 2 y 2 2 y x 3 ( x + y ) 2
Use the initial relationship given on the RHS,
y = 3 ( x + y ) 3 x + y y 2 2 y x 3 ( x + y ) 3 x + y
y = 3 x y 2 x + y y 2 2 y x 3 x y 2 x + y
The rest is just algebra,
y = 3 x y 2 y 2 ( x + y ) 2 y x ( x + y ) 3 x y 2
y = 3 x y 2 x y 2 y 3 2 y x 2 + 2 y 2 x 3 x y 2
y = 2 x y 2 y 3 2 y x 2 x y 2
y = y 2 ( 2 x y ) y x ( 2 x y )
y = y x
as requiblack, et voilà!

EDIT: And just to flesh out hint involving logarithmic differentiation:
3 log x + 6 log y = 9 log ( x + y )
3 x + 6 y y = 9 x + y ( 1 + y )
y ( 6 y 9 x + y ) = 9 x + y 3 x
y ( 6 x 3 y y ( x + y ) ) = 6 x 3 y x ( x + y )
y = y x
A decidedly much quicker route to the answer.
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spatularificw2
Answered 2022-10-10 Author has 1 answers
For
( x + y ) m + n = x m y n
Take logarithm wrt base e ,
( m + n ) ln ( x + y ) = m ln x + n ln y
Now differentiate wrt x
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Please excuse if the formatting of this post is wrong.

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