Chapter 2.7, Problem 22E

To determine

To calculate: The first and second derivative of the function $M\left(t\right)=\frac{t}{1+t}$ for $t>0$ also sketch the graph of the function.

Answer to Problem 22E

The value of first derivative of the function is $M\text{'}\left(t\right)=\frac{1}{{\left(1+t\right)}^2}$ and second derivative is $M\text{'}\text{'}\left(t\right)=\frac{-2}{{\left(1+t\right)}^3}$ . The graph of the function $M\left(t\right)=\frac{t}{1+t}$ for $t>0$ is,

  

Explanation of Solution

Given information:

The function $M\left(t\right)=\frac{t}{1+t}$ .

Formula used:

Let a function g be continuous on closed interval $\left[c,d\right]$ and differentiable on open interval $\left(c,d\right)$ ,

If first derivative of the function is greater than zero that is $g\text{'}\left(x\right)>0$ for every x in $\left(c,d\right)$ , then the function g is increasing on interval $\left[c,d\right]$ .

If first derivative of the function is less than zero that is $g\text{'}\left(x\right)<0$ for every x in $\left(c,d\right)$ , then the function g is decreasing on interval $\left[c,d\right]$ .

If second derivative of the function is greater than zero that is $g\text{'}\text{'}\left(x\right)>0$ for every x in $\left(c,d\right)$ , then the function g is concave up on interval $\left[c,d\right]$ .

If second derivative of the function is less than zero that is $g\text{'}\text{'}\left(x\right)<0$ for every x in $\left(c,d\right)$ , then the function g is concave down on interval $\left[c,d\right]$ .

The point where the graph changes it nature is known as the point of inflection.

Quotient rule of differentiation, $\frac{d}{dx}\left(\frac{f}{g}\right)=\frac{gf\text{'}-fg\text{'}}{g^2}$ .

Calculation:

Consider the provided function $M\left(t\right)=\frac{t}{1+t}$ .

Evaluate the first derivative of the function, apply the quotient rule of differentiation, $\frac{d}{dx}\left(\frac{f}{g}\right)=\frac{gf\text{'}-fg\text{'}}{g^2}$ .

  $\begin{array}{c}\frac{d}{dt}M\left(t\right)=\frac{d}{dt}\left(\frac{t}{1+t}\right)\\ =\frac{\left(1+t\right)\cdot 1-t\left(0+1\right)}{{\left(1+t\right)}^2}\\ =\frac{1}{{\left(1+t\right)}^2}\end{array}$

Evaluate the second derivative of the function, differentiate the first derivative again with respect to x .

  $\frac{d^2}{d{t}^2}M\left(t\right)=\frac{d}{dt}\left(\frac{1}{{\left(1+t\right)}^2}\right)$

Apply the quotient rule of differentiation, $\frac{d}{dx}\left(\frac{f}{g}\right)=\frac{gf\text{'}-fg\text{'}}{g^2}$ .

  $\begin{array}{c}\frac{d^2}{d{t}^2}M\left(t\right)=\frac{d}{dt}\left(\frac{1}{{\left(1+t\right)}^2}\right)\\ =\frac{{\left(1+t\right)}^1\cdot 0-1\cdot 2\left(1+t\right)}{{\left(1+t\right)}^4}\\ =\frac{-2\left(1+t\right)}{{\left(1+t\right)}^3}\\ =\frac{-2}{{\left(1+t\right)}^3}\end{array}$

Recall if first derivative of the function is greater than zero that is $g\text{'}\left(x\right)>0$ for every x in $\left(c,d\right)$ , then the function g is increasing on interval $\left[c,d\right]$ .

If first derivative of the function is less than zero that is $g\text{'}\left(x\right)<0$ for every x in $\left(c,d\right)$ , then the function g is decreasing on interval $\left[c,d\right]$ .

If second derivative of the function is greater than zero that is $g\text{'}\text{'}\left(x\right)>0$ for every x in $\left(c,d\right)$ , then the function g is concave up on interval $\left[c,d\right]$ .

If second derivative of the function is less than zero that is $g\text{'}\text{'}\left(x\right)<0$ for every x in $\left(c,d\right)$ , then the function g is concave down on interval $\left[c,d\right]$ .

To sketch the graph of the function $M\left(t\right)=\frac{t}{1+t}$ follow the steps below,

Observe that first derivative of the function $M\text{'}\left(t\right)=\frac{1}{{\left(1+t\right)}^2}$ is positive for $t>0$ , so $M\left(t\right)=\frac{t}{1+t}$ is an increasing function.

Next observe that second derivative of the function $M\text{'}\text{'}\left(t\right)=\frac{-2}{{\left(1+t\right)}^3}$ is negative for $t>0$ , so $M\left(t\right)=\frac{t}{1+t}$ is concave down always.

Therefore, the graph of the function $M\left(t\right)=\frac{t}{1+t}$ for $t>0$ is provided below,

  

Thus, the value of first derivative of the function is $M\text{'}\left(t\right)=\frac{1}{{\left(1+t\right)}^2}$ and second derivative is $M\text{'}\text{'}\left(t\right)=\frac{-2}{{\left(1+t\right)}^3}$ .