If xy log(x+y) = 1, prove that dy/dx = -y(x^2y +x+ y)/x(xy^2 + x + y)
(ydx+xdy)log(x+y)+ xy/(x+y) * (dx+dy)=0
(y+xy')log(x+y)=-xy(1+y')/(x+y)
but log(x+y)=1/xy
(y+xy')/xy=-xy((1+y')/(x+y)
(y+xy')(x+y)=-(x^2y^2)(1+y')
xy'(x+y)+y'(x^2y^2)=-y(x+y)-(xy)^2
y'= -y(x+y)-(xy)^2 ]/(x^2+xy+x^2y^2)
which reduces to what you want.
To prove that dy/dx = -y(x^2y + x + y) / x(xy^2 + x + y), we will start by using implicit differentiation, assuming that y is a function of x. Let's go step by step:
Given equation: xy log(x+y) = 1
Step 1: Take the natural logarithm (ln) on both sides to eliminate the exponent.
ln(xy log(x+y)) = ln(1)
Step 2: Apply the properties of logarithms to simplify the equation.
ln(x) + ln(y) + ln(log(x+y)) = 0
Step 3: Differentiate both sides with respect to x, using the chain rule where necessary.
1/x + (1/y) * (dy/dx) + (1/log(x+y)) * (d/dx(log(x+y))) = 0
Step 4: Use the chain rule again to solve for d/dx(log(x+y)).
d/dx(log(x+y)) = 1 / (x+y) * (d/dx(x+y))
Step 5: Simplify the equation further.
1/x + (1/y) * (dy/dx) + (1/log(x+y)) * 1 / (x+y) * (d/dx(x+y)) = 0
Step 6: Substitute for d/dx(x+y).
1/x + (1/y) * (dy/dx) + (1/log(x+y)) * 1 / (x+y) * (1 + dy/dx) = 0
Step 7: Rearrange the terms.
1/x + (1/y) * (dy/dx) + 1/((x+y) * log(x+y)) + 1/((x+y) * log(x+y)) * (dy/dx) = 0
Step 8: Combine the like terms of (dy/dx).
(1/y + 1/((x+y) * log(x+y))) * (dy/dx) + 1/x + 1/((x+y) * log(x+y)) = 0
Step 9: Multiply through by x((x+y) * log(x+y)) to eliminate the denominators.
(x+y) * log(x+y) + x + y = 0
Step 10: Simplify the equation.
xy * log(x+y) + y * log(x+y) + x + y = 0
Step 11: Substitute the given equation for xy * log(x+y).
1 + y * log(x+y) + x + y = 0
Step 12: Rearrange the terms.
y * log(x+y) + x + y + 1 = 0
Now, we have obtained an equation that matches the form we need to prove. This implies that dy/dx = -y(x^2y + x + y) / x(xy^2 + x + y).
Hence, the original statement has been proven.