use implicit differentiation to find dy/dx
xlny=y^3=lnx
Can you check the question? There are two equations (equal signs). Could one of them be an operator instead?
To find dy/dx using implicit differentiation, we'll need to differentiate both sides of the equation with respect to x, treating y as a function of x.
Let's start differentiating each term step by step:
1. Differentiating the first term, xlny, with respect to x using the product rule:
d/dx (xlny) = 1 * ln(y) + x * (1/y) * (dy/dx)
2. Differentiating the second term, y^3, with respect to x using the chain rule:
d/dx (y^3) = 3y^2 * (dy/dx)
3. Differentiating the third term, lnx, with respect to x:
d/dx (lnx) = 1/x
Now let's substitute these derivatives back into the equation and solve for dy/dx:
1 * ln(y) + x * (1/y) * (dy/dx) = 3y^2 * (dy/dx) + 1/x
Next, let's rearrange the equation to isolate dy/dx terms on one side:
1 * ln(y) - 3y^2 * (dy/dx) = - x * (1/y) - 1/x
Now, factor out dy/dx:
dy/dx * (x * (3y^2) - 1) = - x/y - 1/x - ln(y)
Divide both sides by (x * 3y^2 - 1):
dy/dx = (-1/x - x/y - ln(y)) / (x * 3y^2 - 1)
And that's our answer for dy/dx using implicit differentiation!