differentiate $\frac {dx}{dt}$ when $x=(1+e^{-2u})^{-0.5}$

Question

I am stuck in what would be simple differentiation:

When $\dot x =\frac {dx}{dt}$ now when $x=(1+e^{-2u})^{-0.5}$ I think I can write: $$\dot x =\frac {d(1+e^{-2u})^{-0.5} }{dt}$$

But can I proceed with:

$$\dot x =- \frac{1}{2}(-2) (1+e^{-2 \dot u})^{-1.5} $$

Answer 1

If $u$ is a function of $t$, by the chain rule

$$\dot x=-0.5(1+e^{-2u})^{-1.5}\ (-2)e^{-2u}\ \dot u.$$

Answer 2

Unless you meant $e^{-2t}$, you'll need implicit differentiation.

Here, we use the chain rule, which states that:

$$x=[f(u)]^n \to \frac {dx}{dt} =n\cdot f'(u)\frac {du}{dt}[f(u)]^{n-1} $$

Answer 3

If $f(y) = \left(1+e^{-2y}\right)^{-0.5}$ (where $y$ is any real number), then you have $$ x(t) = f(u(t)), $$ so by the Chain Rule, $$ x'(t) = f'(u(t))u'(t), $$ i.e., $$ \dot{x} = f'(u)\dot{u}. $$ But the dot notation led you astray. You wrote, in effect, $$ x'(t) = f'(u'(t)) \ \ \text{[Wrong.]} $$ That's not all, though: you also went wrong in differentiating $f$ itself. Making the use of the Chain Rule more explicit, for a second time, let $z$ be any strictly positive real number, and let $g(z) = (1+z)^{-0.5}$, so that $$ f(y) = g\!\left(e^{-2y}\right). $$ Then $$ f'(y) = g'\!\left(e^{-2y}\right)\left(-2e^{-2y}\right), $$ and, although you differentiated $g$ correctly, you lost a factor of $e^{-2y} = e^{-2u(t)}$ along the way.