Expert: Paul Klarreich - 3/20/2007

Question


Given the functions  y = 3x^3 + 6x  and x = 2t^2 + t find
dy/dt.

and calculate dy/dx  if Ln (x+y ) = e^(x/y)


Thanks


Answer
Questioner:   Luciano.Bacco
Category:  Calculus
Private:  No
 
Subject:  Calculus
Question:  

Given the functions  y = 3x^3 + 6x  and x = 2t^2 + t find
dy/dt.

and calculate dy/dx  if Ln (x+y ) = e^(x/y)


Thanks
...................................
Hi, Luciano,

Given  y = 3x^3 + 6x,  x = 2t^2 + t, you will have:

dy/dx = 9x^2 + 6,  and

dx/dt = 4t + 1

Now apply the Chain Rule:

dy   dy dx
-- = -- --
dt   dx dt

dy/dt = (9x^2 + 6)(4t + 1)

But I'm sure you want an answer in terms of t, so substitute x = 2t^2 + t:

dy/dt = (9(2t^2 + t)^2 + 6)(4t + 1)

Multiply out:

dy/dt = (9(4t^4 + 4t^3 + t^2) + 6)(4t + 1)

dy/dt = (36t^4 + 36^3 + 9t^2 + 6)(4t + 1)

and you can multiply out further, if you like.

..................................

Ln (x+y ) = e^(x/y)

An implicit differentiation example.
 1                           y(1) - x dy/dx
-----(1 + dy/dx) =  e^(x/y)[  ---------------  ]
x + y                              y^2

Now I think I would like to clear fractions by multiplying by y^2(x + y):

y^2(1 + dy/dx) = e^xy(x + y)(y - x dy/dx)

Multiply out:

y^2 + y^2 dy/dx = e^xy(x + y)(y)  - e^xy(x + y)(x) dy/dx

Rearrange terms:

y^2 dy/dx + y^2  e^xy(x + y)(x) dy/dx = e^xy(x + y)(y) - y^2

Factor out  dy/dx, because we are solving for it:

(y^2 + y^2 e^xy(x + y)x) dy/dx = e^xy(x + y)(y) - y^2

Divide:
        e^xy(x + y)(y) - y^2
dy/dx = -----------------------
        y^2 + y^2 e^xy(x + y)x

You can probably do a little more, such as:

        y(e^xy(x + y) - y)
dy/dx = -----------------------
        y^2(1 + e^xy(x + y)x)

        e^xy(x + y) - y
dy/dx = -----------------------
        y(1 + e^xy(x + y)x)