Suppose (x^2)*(e^(-2y))= ln(xy)
Find dy/dx by implicit differentiation.
2x(e^(-2y) ) + x^2(e^(-2y) )(-2dy/dx)= 1/(xy) (xdy/dx + y)
2x e^(-2y) -2y x^2 e^(-2y) = (1/y) dy/dx + 1/x
2x e^(-2y) - 1/x = dy/dx (1/y + 2x^2 e(-2y) )
dy/dx = ( 2x e^(-2y) - 1/x ) / ((1/y + 2x^2 e(-2y)) )
still pretty messy looking, I don't know how far you have to take it.
Also, you better check my work. It is always harder to just type it here, rather than do it on paper.
To find dy/dx by implicit differentiation, we will differentiate both sides of the equation with respect to x, treating y as a function of x.
Let's differentiate each term step by step:
For the left side: (x^2)*(e^(-2y))
To differentiate (x^2) with respect to x, we use the power rule. The derivative of (x^2) is 2x.
To differentiate (e^(-2y)) with respect to x, we use the chain rule. The derivative of e^u is (e^u)(du/dx), where du/dx is the derivative of the exponent with respect to x.
So, (e^(-2y)) differentiates to (e^(-2y))*(-2(dy/dx)) using the chain rule.
Therefore, the left side becomes: 2x*(e^(-2y))*(-2(dy/dx)) = -4x*(e^(-2y))*(dy/dx)
For the right side: ln(xy)
To differentiate ln(xy), we use the chain rule. The derivative of ln(u) is (1/u)*(du/dx), where u is the argument of ln.
So, ln(xy) differentiates to (1/(xy))*(d/dx(xy)) using the chain rule.
To find d/dx(xy), we apply the product rule. The derivative of xy with respect to x is (1*y) + (x*dy/dx) = y + x*(dy/dx).
Therefore, the right side becomes: (1/(xy))*(y + x*(dy/dx)) = (y/(xy)) + (x/(xy))*(dy/dx) = (y/x) + (dy/dx)
Now, equating the left and right sides:
-4x*(e^(-2y))*(dy/dx) = (y/x) + (dy/dx)
Let's isolate dy/dx on one side by moving the terms involving dy/dx to the left:
-4x*(e^(-2y))*(dy/dx) - (dy/dx) = (y/x)
Factoring out dy/dx:
[ -4x*(e^(-2y)) - 1 ] * (dy/dx) = (y/x)
Finally, divide both sides by [-4x*(e^(-2y)) - 1] to solve for dy/dx:
(dy/dx) = (y/x) / [-4x*(e^(-2y)) - 1]
This is the expression for dy/dx obtained through implicit differentiation.