I believe this is e rule but i'm not sure.. help?

Question

Find y′ if x^(3y) = y^(2x).

 

1. y′=  3y^2(ln x − 1)
             3y ln x + 2x

 

2. y′ =  y^2(ln x − 1)

            3y ln x − 2x

 

3. y′ =  y(2x ln y − 3y)
            x(3y ln x − 2x)

 

4. y′ = 3(y^2)(ln x + 1)
              3y ln x − 2x
5. y′ = y(3(x^2)ln y − 2y)
           x(2(y^2)ln x − 3x)

Answer 1

x^3y = y^2x

 

Log both sides of the equation and bring any exponents down as the coefficient of the Log.

 

3yLn(x) = 2xLn(y)

 

 

Now we can implicit differentiate.

 

3y'Ln(x) + (3y/x) = 2Ln(y) + (2xy'/y)

 

 

Multiply both sides of the equation by x.

 

3xy'Ln(x) + 3y = 2xLn(y) + (2x²y'/y)

 

 

Multiply both sides of the equation by y.

 

3xyy'Ln(x) + 3y² = 2xyLn(y) + 2x²y'

 

 

Isolate all the y'-terms.

 

3xyy'Ln(x) - 2x²y' = 2xyLn(y) - 3y²

 

 

Next, just factor out y' on the left side. 

 

y' [3xyLn(x) - 2x²] = 2xyLn(y) - 3y²

 

 

Finally, divide both sides of the equation by the coefficient of y'.

Answer 2

You can get y' as a function of x and y.

 

x^3y = y^2x

 

Take ln() on both sides.

 

3y lnx = 2x lny

 

Differentiate both sides.

 

3(y' ln x + y/x) = 2(lny +xy'/y)

 

Solve for y'.

 

y'(3 lnx - 2x/y) = 2 lny - 3y/x

 

y' = (2 lny - 3y/x) / (3 lnx - 2x/y)