Question

How do you find the derivative of `y = ln[(x+3)/x^2)]`?

Answer 1

`dy/dx = -(x +6)/(x(x +3))`

Explanation:

We let `y = lnu` and `u = (x + 3)/x^2`

Then `dy/(du) = 1/u` and `(du)/dx = (1(x^2) - 2x(x + 3))/(x^2)^2 = (x^2 - 2x^2 - 6x)/x^4 = (-x(x + 6))/x^4 = -(x + 6)/x^3`

The chain rule states that `color(red)(dy/dx= dy/(du) xx (du)/dx`.

`dy/dx = 1/u xx -(x + 6)/x^3`

`dy/dx= (-(x + 6)/x^3)/((x + 3)/x^2)`

`dy/dx= -(x + 6)/x^3 xx x^2/(x + 3)`

`dy/dx = -(x +6)/(x(x +3))`

Hopefully this helps!

Answer 2

We can also rewrite this the logarithm rules `log(a/b)=log(a)-log(b)` and `log(c^d)=dlog(c)`.

`y=ln((x+3)/x^2)=ln(x+3)-ln(x^2)`

`color(white)(y=ln((x+3)/x^2))=ln(x+3)-2ln(x)`

We should know that `d/dxln(x)=1/x`. Applying the chain rule to this shows us that `d/dxln(u)=1/u*(du)/dx`.

Then:

`dy/dx=1/(x+3)*d/dx(x+3)-2(1/x)`

Since `d/dx(x+3)=1`:

`dy/dx=1/(x+3)-2/x`

Finding a common denominator:

`dy/dx=(x-2(x+3))/(x(x+3))`

`color(white)(dy/dx)=(-x-6)/(x(x+3))`

`color(white)(dy/dx)=-(x+6)/(x(x+3))`