Why does squaring an equation before differentiation give a different result?

Question

Consider $t = \sqrt{x -1}$.

Suppose I want to find the velocity $\frac{dx}{dt}$ at $x =1$. I proceed with two different approaches:

$1st$ APPROACH:

I differentiate both sides with respect to $t$:

\begin{equation} \begin{aligned} 1 &= \frac{\frac{dx}{dt}}{2\sqrt{x- 1}} \\ \implies \frac{dx}{dt} &= 2\sqrt{x- 1} \end{aligned} \end{equation} At $x=1$, we get $\frac{dx}{dt} = 0.$

---

$2nd$ APPROACH:

I square both sides and get $t^2 =x - 1$.

Differentiating both sides, we get:

$2t = \frac{dx}{dt}$

How do these two approaches generate two different results?

Answer 1

Both approaches are valid, but they are still giving the same result. You can substitute $t = \sqrt{x-1}$ into the equation $2t = \frac{dx}{dt}$ to see that the second approach gives the same formula as the first one.

Answer 2

They do not give different results. In your second attempt, you get $\frac{dx}{dt}=2t$. But when $x=1$ we have $t=0$, so this evaluates to $$ \frac{dx}{dt}\Bigr|_{t=0}=2\cdot 0 =0 $$ just as in your first effort.