Is $dv$ only approximate of $(dv/dx)dx$?

Question

I've tried to solve this problem:

If $V=2x^{3}$ what is the approximate percentage change in $V$ when $x$ changes by 2%?

My solution is:

Since $x$ changes by 2% then $dx=0.02x$. $\frac{dv}{dx} = 6x^{2}$. I think that: $$dv=\frac{dv}{dx}dx$$

So $dv=6x^{2}*0.02x=0.12x$

So the percentage change in $V$ is $\frac{dV}{V}=\frac{0.12x}{2x^{3}}=0.06=6\%$ I believe that is correct.

However, the answer says that $$dv\approx \frac{dv}{dx}dx$$ Notice the approximate sign. I don't know why this should be an approximation. Could anyone explain where is the part we're approximating? I tried to compare this to "normal" functions where e.g. if we have a value of $x$ and the gradient at this point we can estimate the value of $f(x)$ at for example $x=x+1$. But in this case they are infinitesimal values so $\frac{dv}{dx}$ shouldn't change, right?

Answer 1

Let $x' = x+0.02x$, then $V' = 2(x+0.02x)^3 = 2(1.02x)^3$.

The percentage change in $V$ is $$\begin{align*} \frac{V' - V}{V}&= \frac{2(1.02x)^3-2x^3}{2x^3}\\ &= 1.02^3-1\\ &= 6.1208\% \end{align*}$$

For a negative $2\%$ change in $x$, the percentage change in $V$ is similar: $$\frac{V'-V}V = 0.98^3-1 = -5.8808\%$$

Answer 2

The answer you cite is using the symbols $dV$ and $dx$ in two senses. What it means is something more like

$$\Delta V \approx \frac{dV}{dx} \Delta x$$ where $\Delta V$ is the change in $V$ as an ordinary real number. By definition

$$\frac{dV}{dx} = \lim_{\Delta x \to 0} \frac{\Delta V}{\Delta x}$$

and hence it must be the case that we can make $\displaystyle\frac{\Delta V}{\Delta x}$ as close as we like to $\displaystyle\frac{dV}{dx}$ by making $\Delta x$ sufficiently small.

Now, the percentage change is $\displaystyle \frac{\Delta x}{x}$ and $\displaystyle \frac{\Delta V}{V}$. For your problem at hand it is useful to write the relation $V = 2x^3$ as

$$\ln V = \ln(2x^3) = \ln 2 + 3\ln x$$

Useful, because when we now differentiate both sides with respect to $x$,

$$\frac 1V \frac{dV}{dx} = \frac 3x$$

or

$$\frac 1V \frac{\Delta V}{\Delta x} \approx \frac 3x \quad\text{ by the logic above }$$ and thus

$$\frac{\Delta V}{V} \approx 3\frac{\Delta x}{x} \, .$$

Hence as 2% change in $x$ is (approximately) equivalent to a 6% change in $V$.