Chapter 2.9, Problem 29E

To determine

To calculate: The derivative of the inverse of the function, $q\left(x\right)=x+x^2$ for $x\ge 0$ by two ways first evaluate the inverse then the derivative second by derivative of inverse.

Answer to Problem 29E

The derivative when inverse is evaluated first then its derivative is $\left(q^{-1}\right)\text{'}\left(y\right)=\frac{1}{\sqrt{1+4y}}$ and derivative of inverse of function is $q\text{'}\left[q^{-1}\left(y\right)\right]=\sqrt{1+4y}$ .

Explanation of Solution

Given information:

The function, $q\left(x\right)=x+x^2$ for $x\ge 0$ .

Formula used:

Let a function g is differentiable and has inverse denoted by $g^{-1}$ such that $g\text{'}\left[g^{-1}\left(x\right)\right]\ne 0$ , then the derivative of the function $g^{-1}\left(x\right)$ is, $\left(g^{-1}\right)\text{'}\left(x\right)=\frac{1}{g\text{'}\left[g^{-1}\left(x\right)\right]}$ , where $g\text{'}$ denotes the derivative of function g .

Calculation:

Consider the provided function, $q\left(x\right)=x+x^2$ for $x\ge 0$ .

First evaluate the inverse of the function, $q\left(x\right)=x+x^2$ for $x\ge 0$ .

Let $y=x+x^2$ , solve the equation for x in terms of y .

Rewrite the equation and apply the quadratic formula,

  $\begin{array}{c}y=x+x^2\\ x^2+x-y=0\\ x=\frac{-1\pm \sqrt{1+4y}}{2}\end{array}$

Since, it is provided that domain of the function is $x\ge 0$ , so $x=\frac{-1+\sqrt{1+4y}}{2}$ is considered.

Therefore, inverse of the function is $q^{-1}\left(y\right)=\frac{-1+\sqrt{1+4y}}{2}$ .

Next derivative of the inverse evaluated above is, apply the chain rule of differentiation.

Let $G\left(y\right)=f\left(g\left(y\right)\right)$ , where $g\left(y\right)=\sqrt{1+4y}$ and $f\left(g\right)=\frac{-1+g}{2}$ .

Differentiate the individual functions, apply power rule of differentiation $\frac{d}{dx}{x}^n=n{x}^{n-1}$ .

  $\begin{array}{c}g\left(y\right)=\sqrt{1+4y}\\ g\left(y\right)={\left(1+4y\right)}^{1/2}\\ g\text{'}\left(y\right)=\frac{1}{2}{\left(1+4y\right)}^{-1/2}\cdot 4\\ g\text{'}\left(y\right)=\frac{2}{\sqrt{1+4y}}\end{array}$

And

  $f\text{'}\left(g\right)=\frac{1}{2}$

Therefore, derivative of $G\left(y\right)=f\left(g\left(y\right)\right)$ is $G\text{'}\left(y\right)=f\text{'}\left(g\right)\cdot g\left(y\right)$ , that is,

  $\begin{array}{c}G\text{'}\left(y\right)=f\text{'}\left(g\right)\cdot g\text{'}\left(y\right)\\ =\frac{1}{2}\cdot \frac{2}{\sqrt{1+4y}}\\ =\frac{1}{\sqrt{1+4y}}\end{array}$

Therefore, derivative when inverse is evaluated first then its derivative is $\left(q^{-1}\right)\text{'}\left(y\right)=\frac{1}{\sqrt{1+4y}}$ .

The derivative of provided function $q\left(x\right)=x+x^2$ is, apply the power rule of differentiation $\frac{d}{dx}{x}^n=n{x}^{n-1}$ .

That is,

  $q\text{'}\left(x\right)=1+2x$ .

It is known that polynomial functions are always differentiable.

Recall if a function g is differentiable and has inverse denoted by $g^{-1}$ such that $g\text{'}\left[g^{-1}\left(x\right)\right]\ne 0$ , then the derivative of the function $g^{-1}\left(x\right)$ is, $\left(g^{-1}\right)\text{'}\left(x\right)=\frac{1}{g\text{'}\left[g^{-1}\left(x\right)\right]}$ , where $g\text{'}$ denotes the derivative of function g .

Apply it,

  $\begin{array}{c}\left(q^{-1}\right)\text{'}\left(y\right)=\frac{1}{q\text{'}\left[q^{-1}\left(y\right)\right]}\\ q\text{'}\left[q^{-1}\left(y\right)\right]=\frac{1}{\left(q^{-1}\right)\text{'}\left(y\right)}\\ =\frac{1}{\frac{1}{\sqrt{1+4y}}}\\ =\sqrt{1+4y}\end{array}$

Therefore, the derivative of inverse of function is $q\text{'}\left[q^{-1}\left(y\right)\right]=\sqrt{1+4y}$ .

Thus, derivative when inverse is evaluated first then its derivative is $\left(q^{-1}\right)\text{'}\left(y\right)=\frac{1}{\sqrt{1+4y}}$ and derivative of inverse of function is $q\text{'}\left[q^{-1}\left(y\right)\right]=\sqrt{1+4y}$ .