Question

What is the derivative of `y = ln((2x)/(x+3))`?

Answer 1

The derivative can be found using the chain rule.

However, we must first determine the derivative for `y = lnx`. The answer: `y' = 1/x`. The reason for this is too complicated and might mix you up.

Now, by function composition, we have:

`y = ln(u)`

`u = (2x)/(x + 3)`

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

We must differentiate both functions.

`y' = 1/u`

By the quotient rule:

`u' = ((2(x + 3)) - (2x)(1))/(x + 3)^2`

`u' = (2x + 6 - 2x)/(x+ 3)^2`

`u' = 6/(x + 3)^2`

`:.dy/dx = dy/(du) xx (du)/dx`

`dy/dx = 1/u xx 6/(x + 3)^2`

`dy/dx = 1/((2x)/(x + 3)) xx 6/(x + 3)^2`

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

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

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

The derivative of `y = ln((2x)/(x + 3))3` is `dy/dx = 6/(x^2 + 3x)`

Practice exercises:

`1.` Differentiate `y = root(3)(2x + 7)` by the chain rule.

`2.` Differentiate `y = ln(2x^3 + 5x^2 - 4x + 1)` using the chain rule.

`3.` Differentiate `y = ln((x^2 + 5x)/(x^3))` by the chain rule.

Note: The power rule can be helpful when differentiating some of the functions above. This rule states that for a function `y = ax^n`, the derivative is given by `y = a xx n xx x^(n - 1)`. This can be used in addition to the sum and difference rule. This rule states that for a function `y = f(x) + g(x)`, the derivative is given by `y = f'(x) + g'(x)`.

Hopefully this helps, and good luck!

Answer 2

`y^'=3/(x^2+3x)`

Explanation:

Another approach is to first simplify the natural logarithm expression using the rule `log(A/B)=log(A)-log(B)`. Thus:

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

We can even simplify `ln(2x)` as `ln(2)+ln(x)` using the other rule `log(A*B)=log(A)+log(B)`:

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

When differentiating this, keep a couple things in mind. Primarily, `ln(2)` is just a constant, so its derivative is `0`. Secondly, the derivative of `ln(x)` is `1/x`. This will also be necessary for using the chain rule while differentiating `ln(x+3)`.

`y^'=1/x-1/(x+3)*d/dx(x+3)`

`y^'=1/x-1/(x+3)`

We can combine the fractions:

`y^'=(x+3)/(x(x+3))-x/(x(x+3))`

`y^'=3/(x^2+3x)`