Find dy/dx in terms of x and y if xlny+y^6= (9)lnx

x lny+ y^6=9lnx

lny +xy'/y+6y^5 y'=9/x

solve for y'

I solved and I got: (9-lny)/(x((x/y)+6y^5))

But that's not the answer...

To find dy/dx, we need to differentiate both sides of the equation with respect to x.

Let's differentiate each term step-by-step using the rules of differentiation.

Starting with the left side of the equation:
d/dx(xlny) + d/dx(y^6) = d/dx((9)lnx)

To differentiate xlny:
Using the product rule, the derivative of xlny with respect to x is given by:
(d/dx(x))(lny) + x(d/dx(lny))

The derivative of x with respect to x is 1, and the derivative of lny with respect to x can be found using the chain rule as (1/y)(dy/dx).

So, the first term becomes:
1(lny) + x(1/y)(dy/dx)

Now, let's differentiate y^6:
Using the chain rule, the derivative of y^6 with respect to x is:
(6y^5)(dy/dx)

Finally, the derivative of (9)lnx with respect to x is simply 0, since the derivative of a constant (9) is zero.

Putting it all together, we have:
1(lny) + x(1/y)(dy/dx) + (6y^5)(dy/dx) = 0

Rearranging the terms to isolate dy/dx:
(x/y)(dy/dx) + (6y^5)(dy/dx) = -1(lny)

Factoring out dy/dx:
[(x/y) + (6y^5)](dy/dx) = -1(lny)

Finally, dividing both sides by [(x/y) + (6y^5)] will yield the derivative dy/dx in terms of x and y:

dy/dx = -1(lny)/[(x/y) + (6y^5)]

Therefore, the derivative dy/dx in terms of x and y is -1(lny)/[(x/y) + (6y^5)].