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)].