differentiate y = ln(x) / (x+2)^2?

Question

Can i take exp(y) = exp(x(x+2)^-2) and find

exp(y)y' = -2x(x+2)^-3
y' = -2x / (x+2)^2

is there something wrong here?

Answer 1

I can't tell what you have in mind, but your derivative is wrong.
Write this as y = (ln x)*(x + 2)^(-2) and differentiate by the product rule: y' = (ln x)*(-2)*(x+2)^(-3) + (1/x)*(x + 2)^(-2). You can clean it up a little: y' = (-2)*(ln x)/(x + 2)^3 + 1/(x*(x + 2)^2).

Answer 2

given, y = ln(x) / (x+2)^2

so, y' = ( (x+2)^2 . (1/x) - ln(x) . 2(x+2) ) / ( x+2)^4
or, y' = ( (x+2)^2 / x(x+2)^4 ) - ( ln(x) . 2(x+2) / (x+2)^4 )
or, y' = ( 1 / x(x+2)^2 ) - ( 2ln(x) / (x+2)^3 )

Answer 3

y = ln(x) / (x+2)²
y = (lnx) * (x+2) -²
dy/dx = (lnx)(-2)(x+2)^-3(1) + (x+2) -²(1/x)
dy/dx = (-2)(lnx)(x+2)^-3 + 1/(x+2)² (1/x)
dy/dx = -2lnx / (x+2)^3 + 1/(x+2)² (1/x)
dy/dx = -2lnx / (x+2)^3 + (x + x + 2)/ (x + 2)(x)
dy/dx = -2lnx / (x+2)^3 + (2x + 2)/ (x² + 2x)