Sometimes it becomes tedious to derive functions that are the result of exponential functions, such is the case of the following example where we are asked to derive the function:
When seeing the exercise the first thing that occurs to us is to apply derivative of a quotient, and that is correct. It should also be taken into account that the denominator is a product, so at some point we will have to derive dof a product and we will also need to apply the chain rule.

If we follow this path, we will surely reach the desired result, but there is a more expeditious way to reach that result and it is the logarithmic differentiation. Specifically, we will use natural logarithm function.

Process:

It begins by applying natural logarithm to both sides of equality:

Then properties of the logarithms are applied, which are also valid for the natural logarithm function.

The first property we are going to use is this: Ln a / b = Lna-Lnb (Logarithm of a quotient)

Our expression looks like this:

Then the following property is applied: Ln ab = Ln a + ln b (Logarithm of a product)
Removing the brackets looks like this:

The following property that we will use is the following: Ln a^n = n Lna (Logarithm of a power)

Finally we apply implicit differentiation and the derivative of the function natural logarithm that is given by the following formula: Dx (Ln U) = (1 / U) U'

Working with the lowest common multiple between the denominators we have:

Reducing similar terms in the numerator:

Solving for y ', we have:

Substituting "y" for its value:

Applying properties of the exponents, we finally have the desired result.