Question

Find the values of several derivatives of${\mathit{e}}^{\mathbf{-}\mathbf{1}\mathbf{/}{\mathbf{t}}^{\mathbf{2}}}$at t = 0. Hint:Calculate a few derivatives (as functions of t); then make the substitution$\mathit{x}\mathbf{=}\mathbf{1}\mathbf{/}{\mathit{t}}^{\mathbf{2}}$, and use the result of Problem 24(f) or 25.

Short answer

The solution is$\underset{x\to \infty }{lim}\frac{d^2}{d{t}^2}\left(e^{f\left(t\right)}\right)=0$. t

Step-by-step solution

Given information

A function$e^{-1/t^2}$is given.

Definition of Maclaurin Series.

The definition of the Maclaurin series used in this solution is given as:

${\mathit{e}}^{\mathbf{x}}\mathbf{=}\mathbf{1}\mathbf{+}\mathit{x}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{2}\mathbf{!}}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{3}}}{\mathbf{3}\mathbf{!}}\mathbf{+}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{n}}}{\mathbf{n}\mathbf{!}}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{n}}{\left(n+1\right)}\mathbf{+}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}$

Perform differentiation several times and make appropriate substitutions

Assume$f\left(t\right)=-\frac{1}{t^2}$.

This implies $e^{-1/t^2}=e^{f\left(t\right)}$.

$$\begin{gathered} \frac{d}{dt}\left(e^{f\left(t\right)}\right)=f\text{'}\left(t\right)e^{f\left(t\right)} \\ \frac{d^2}{d{t}^2}\left(e^{f\left(t\right)}\right)=\frac{d}{dt}f\text{'}\left(t\right)e^{f\left(t\right)} \\ =\left(f\text{'}\left(t\right)\right)e^{f\left(t\right)}+f"\left(t\right)e^{f\left(t\right)} \\ \frac{d^3}{d{t}^3}\left(e^{f\left(t\right)}\right)=\frac{d}{dt}\left(\left(f\text{'}\left(t\right)\right)e^{f\left(t\right)}+f"\left(t\right)e^{f\left(t\right)}\right) \\ =2f\text{'}\left(t\right)f"\left(t\right)e^{f\left(t\right)}+{\left(f\text{'}\left(t\right)\right)}^3{e}^{f\left(t\right)}+f\text{'}\text{'}\text{'}\left(t\right)e^{f\left(t\right)}+f\text{'}\left(t\right)f\text{'}\text{'}\left(t\right)e^{f\left(t\right)} \end{gathered}$$

Note the values of derivatives of all orders.

Derivatives of all the orders are calculated.

$$\begin{gathered} f\left(t\right)=-\frac{1}{t^2} \\ f\text{'}\left(t\right)=\frac{2}{t^3} \\ f"\left(t\right)=-\frac{6}{t^4} \\ f\text{'}\text{'}\text{'}\left(t\right)=\frac{24}{t^5} \end{gathered}$$

Substitute the values of derivatives in the equation.

Substitute the derivates back in the equation obtained.

$$\begin{gathered} \frac{d}{dt}\left(e^{f\left(t\right)}\right)=\frac{2}{t^3}{e}^{-1/t^2} \\ \frac{d^2}{d{t}^2}\left(e^{f\left(t\right)}\right)=\frac{4}{t^6}{e}^{-1/t^2}-\frac{6}{t^4}{e}^{-1/t^2} \\ \frac{d^3}{d{t}^3}\left(e^{f\left(t\right)}\right)=2f\text{'}\left(t\right)f"\left(t\right)e^{f\left(t\right)}+{\left(f\text{'}\left(t\right)\right)}^3{e}^{f\left(t\right)}+f\text{'}\text{'}\text{'}\left(t\right)e^{f\left(t\right)}+f\text{'}\left(t\right)f\text{'}\text{'}\left(t\right)e^{f\left(t\right)} \\ \frac{d^3}{d{t}^3}\left(e^{f\left(t\right)}\right)=\frac{24e^{\left({}^{-1/t^2}\right)}{t}^4-36e^{\left({}^{-1/t^2}\right)}{t}^2+8e^{\left({}^{-1/t^2}\right)}}{t^9} \end{gathered}$$

It is known that $\underset{x\to \infty }{lim}\frac{d^2}{d{t}^2}\left(e^{f\left(t\right)}\right)=0$.

Continue evaluation.

$$\begin{gathered} \underset{x\to \infty }{lim}\frac{d^2}{d{t}^2}\left(e^{f\left(t\right)}\right)=\underset{x\to \infty }{lim}4x^3{e}^x-\underset{x\to \infty }{lim}6x^2{e}^{-x} \\ \underset{x\to \infty }{lim}\frac{d^2}{d{t}^2}\left(e^{f\left(t\right)}\right)=0-0 \\ =0 \end{gathered}$$

Thus, the solution is $\underset{x\to \infty }{lim}\frac{d^2}{d{t}^2}\left(e^{f\left(t\right)}\right)=0$.