Differentiation with $e$

Question

I'm attempting to understand the sign of the result of this differentiation :

$$ \frac{d}{dt}(1 - e{^{-t^3})} = 0-(-3t^2)e^{-t^3} = 3t^2e^{-t^3} $$

If $e^{-t^3}$ becomes $e^{t^3}$

then is this correct differentiation :

$$ \frac{d}{dt}(1 - e{^{t^3})} = 0-(3t^2)e^{t^3} = -3t^2e^{t^3} $$

Answer 1

Yes that's correct since we are using that by chain rule

$$f(x)=e^{g(x)}\implies f'(x)=g'(x)e^{g(x)}$$

note also that by intuition as $t \to \infty$

• $1 - e^{-t^3}\to 1$ is strictly increasing and indeed $\frac{d}{dt}(1 - e{^{-t^3})}>0$

• $1 - e^{t^3}\to -\infty$ is strictly decreasing and indeed $\frac{d}{dt}(1 - e{^{t^3})}<0$

Answer 2

Yup, all that is correct.

The reason for the sign changing comes from the chain rule. Since we are considering $e^{-t^3}$ in the first one, and the derivative of the exponential is itself, the derivative would be given by

$$\frac{d}{dt} e^{-t^3} = (-t^3)' e^{-t^3}$$

and, similarly, for the positive exponent,

$$\frac{d}{dt} e^{t^3} = (t^3)' e^{t^3}$$

Answer 3

It is :

$$(1-e^{-t^3})' = (1') + (-e^{-t^3})' =0 +(-(-t^3)'e^{-t^3}) = 3t^2e^{-t^3}$$

Similarily follows for your second differentiation as well.

Both are indeed correct.

Above it is used that $(f+g)' = f' + g'$ and that $(e^{f(x)})' = f'(x)e^{f(x)}$, of course if $f$ and $g$ are differentiable.

Answer 4

Both results are special cases of $$\frac{d}{dt}(1-e^f)=\frac{d}{dt}(-e^f)=-\frac{d}{dt}e^f=-\frac{df}{dt}e^f,$$where for constant $c$ the first $=$ uses $\frac{dc}{dt}=0$, the second uses $\frac{d}{dt}(cg)=c\frac{dg}{dt}$, and the third uses the chain rule and $\frac{d}{df}e^f=e^f$. Your first example also uses $-(-x)=x$.

Answer 5

$\dfrac{d}{dt}e^{g(x)}=e^{g(x)}\dfrac{d}{dt} g(x)$