Question

Q. 2

Construct examples of the thing(s) described in the following. Try to find examples that are different than any in the reading.

(a) A function that is its own fourth derivative.

(b) A function whose domain is larger than the domain of its derivative.

(c) Three non-logarithmic functions that are transcendental, but whose derivatives are algebraic.

Short answer

(a) the function $f\left(x\right)=\sin x$is its own fourth derivative.

(b) the domain of the function $f\left(x\right)={\sin }^{-1}x$is larger than the domain of its derivative

(c) transcendental functions whose derivatives are algebraic are following.

$$\begin{gathered} f\left(x\right)={\sin }^{-1}x \\ f\left(x\right)={\tan }^{-1}x \\ f\left(x\right)=se{c}^{-1}x \end{gathered}$$

Step-by-step solution

Step 1. Given information.

(a) A function is its own fourth derivative.

(b) Domain of a function is larger than the domain of its derivative.

(c) derivatives of transcendental non-logarithmic functions are algebraic.

Step 2. Part (a)

Consider a function $f\left(x\right)=\sin x.$

Find the fourth derivative.

role="math" localid="1648764216410" $$\begin{gathered} f\text{'}\left(x\right)=\frac{d}{dx}\sin x \\ f\text{'}\left(x\right)=\cos x \\ f\text{'}\text{'}\left(x\right)=\frac{d}{dx}\cos x \\ f\text{'}\text{'}\left(x\right)=-\sin x \\ f\text{'}\text{'}\text{'}\left(x\right)=\frac{d}{dx}\left(-\sin x\right) \\ f\text{'}\text{'}\text{'}\left(x\right)=-\cos x \\ f\text{'}\text{'}\text{'}\text{'}\left(x\right)=\frac{d}{dx}\left(-\cos x\right) \\ f\text{'}\text{'}\text{'}\text{'}\left(x\right)=-\left(-\sin x\right)=\sin x \end{gathered}$$

So fourth derivative of the function$f\left(x\right)=\sin x$is$\sin x.$

Step 3. Part (b)

Consider a function $f\left(x\right)={\sin }^{-1}x.$

The domain of the function $f\left(x\right)={\sin }^{-1}x$is $\left[-1,1\right].$

Derivative of the function.

$$\begin{gathered} f\text{'}\left(x\right)=\frac{d}{dx}{\sin }^{-1}x \\ f\text{'}\left(x\right)=\frac{1}{\sqrt{1-x^2}} \end{gathered}$$

Domain of role="math" localid="1648764574229" $f\text{'}\left(x\right)=\frac{1}{\sqrt{1-x^2}}$is $\left(-1,1\right).$

$\left[-1,1\right]>\left(-1,1\right)$

So the domain of the function role="math" localid="1648764542847" $f\left(x\right)={\sin }^{-1}x$is larger than the domain of its derivative $f\text{'}\left(x\right)=\frac{1}{\sqrt{1-x^2}}.$

Step 4. Part (c)

Consider a transcendental function $f\left(x\right)={\sin }^{-1}x.$

Derivative of function.

$$\begin{gathered} f\text{'}\left(x\right)=\frac{d}{dx}{\sin }^{-1}x \\ f\text{'}\left(x\right)=\frac{1}{\sqrt{1-x^2}} \end{gathered}$$

Consider a transcendental function$f\left(x\right)={\tan }^{-1}x.$

Derivative of function.

$$\begin{gathered} f\text{'}\left(x\right)=\frac{d}{dx}{\tan }^{-1}x \\ f\text{'}\left(x\right)=\frac{1}{1+x^2} \end{gathered}$$

Consider a transcendental function $f\left(x\right)=se{c}^{-1}x.$

Derivative of function.

$$\begin{gathered} f\text{'}\left(x\right)=\frac{d}{dx}se{c}^{-1}x \\ f\text{'}\left(x\right)=\frac{1}{\left|x\right|\sqrt{x^2-1}} \end{gathered}$$

Derivatives of all functions are algebraic.