# How to interpret relative sensitivity function calculated from first derivative in local sensitivity analysis?

I am trying to do sensitivity analysis on a given function by calculating the 1st-order (partial) derivatives. In a simple case let's say the function is $$y=ax^2$$.

Then the 1st-order derivative is: $$\frac{dy}{dx}=2ax$$ (1)

The relative sensitivity indices is calculated as $$\frac{dy}{y}/\frac{dx}{x}=2$$. (2)

My question is how to interpret this relative sensitivity indices? I have read that it can be interpreted as: 2% changes in $$y$$ if 1% changes in $$x$$. But if we think of increasing $$x$$ 1% to have $$x'=1.01x$$ and $$y'=a(1.01x)^2$$, so it is obviously $$(y'-y)/y\neq2(x'-x)/x$$ as we see in formula (2). Can someone give more insights on the meaning of the relative sensitivity indices and in what case it is more informative than the raw first derivate $$dy$$?

thanks