How to use differentials to estimate the relativistic increase in mass from at 90% speed of light to 92% the question below is from a section on differentials in an old calculus text I am teaching myself with (Calculus, Varberg & Purcell, 6th edition.)
The text gives an answer (9.47%) but does not explain how it is derived. That is what I wish to learn. I understand how to calculate the answer using the equation, but not how to estimate it using differentials.
I believe dv = .02, and I have to multiply that by the derivative of the equation below to find dm. But I am confused by the c^2. (which is in fact a constant, further confounding me.)  Must I first express v in terms of c? (for example .9c^2 / c^2, in order to differentiate?  Or possibly I should try implicit differentiation?  Here's the question- any help would be greatly appreciated.  thank you!
"Einstein's Special Theory of Relativity says that the mass m of an object moving at a velocity v is given by the formula:
m = m0 (1- v^2/c^2)^(-1/2)
Here m0 is the rest mass (mass at velocity 0) and c is the speed of light. Use differentials to estimate the percentage increase of the object as its velocity increases from v=0.90c to 0.92c."
 A: By definition of derivative, if $f:\Bbb R\to\Bbb R$ is differentiable at $x$, then: $$f(x+h)-f(x)=f’(x)\cdot h+o(|h|)$$
If you don’t know what the little-$o$ means, don’t worry - it just means a “small” error, which gives rise to the following.
What’s useful to you is the resulting (and very common) estimate by differentials:
$$\Delta f=f(x+h)-f(x)\approx f’(x)\cdot h=f’\cdot\Delta x$$
When $\Delta x$ is small.
So we estimate the percentage change in mass to be: $$100\cdot\frac{\Delta m}{m}\approx100\cdot\frac{f’(v)}{f(v)}\Delta v$$Where $m=f(v)$ is Einstein’s relativity formula. The derivative of it is:
$$f’(v)=\frac{v}{c^2}m_0\left(1-\frac{v^2}{c^2}\right)^{-3/2}$$
Dividing this by $f(v)$ gives:
$$\frac{f’(v)}{f(v)}=\frac{v}{c^2}\left(1-\frac{v^2}{c^2}\right)^{-1}$$
For this particular estimate, $v=0.9c$ and $\Delta v=0.02c$.
The percentage change in $m$ is then, by substituting into the above equation, roughly:
$$\begin{align}100\cdot0.02c\cdot\frac{0.9c}{c^2}\left(1-\frac{0.81c^2}{c^2}\right)^{-1}&=\frac{2c\cdot0.9}{c}(1-0.81)^{-1}\\&=1.8\cdot(0.19)^{-1}\\&\approx9.47368\end{align}$$
Precisely how accurate this estimate is, I am unsure, but it is a common theme in physics. More generally you might be interested in Taylor approximations; it would have been better if I used the estimation: $$\Delta f\approx f’\cdot\Delta x+\frac{1}{2}f’’\cdot(\Delta x)^2$$
But that would have been more than the question required.
