Chapter 2.5, Problem 50E

a.

To determine

To Find: The derivatives of $\sin x$ and $\cos x$ if $x$ is measured in degrees instead of radians.

Answer to Problem 50E

Its limit is $\frac{\pi }{180}$ as this is the convergence factor from degrees to radians.

Explanation of Solution

Given:

With the grapher in degree mode, graph $f\left(h\right)=\frac{\sin h}{h}$ and estimate $\underset{h\to 0}{\lim }f\left(h\right)$ .

Compare the estimation with $\frac{\pi }{180}$ .

Find the reason if any to believe the limit should be $\frac{\pi }{180}$ .

Graph of $f\left(h\right)=\frac{\sin h}{h}$ in degree mode:

  

The value of $f\left(h\right)$ approaches approximately $0.01745$ when numbers near $0$ are plugged in.

The value of $180$ calculated with a calculator is $0.01745$ , indicating that the values are the same.

To Convert degrees to radians and vice versa:

  $h^{°}=\frac{\pi }{180}h\text{ radians }$

Thus, the limit can be:

  $\begin{array}{c}\underset{h\to 0}{\lim }\frac{\sin h^{°}}{h}=\underset{h\to 0}{\lim }\frac{\sin \left(\frac{\pi }{180}h\right)}{h}\\ =\frac{\pi }{180}\end{array}$

b.

To determine

To Estimate: The given function by setting the grapher in degree mode.

Answer to Problem 50E

The value of the function $\underset{h\to 0}{\lim }\frac{\cos h-1}{h}$ is $0$ according to the graph.

Explanation of Solution

Given:

With the grapher in degree mode, graph $\underset{h\to 0}{\lim }\frac{\cos h-1}{h}$

Graph of the function $\frac{\cos h-1}{h}$:

  

It can be seen from the graph, that the limit as $h$ approaches to $0$ is $0$ .

Thus,

  $\underset{h\to 0}{\lim }\frac{\cos h-1}{h}=0$

c.

To determine

To Find: The formula that obtained for the derivative.

Answer to Problem 50E

The formula that obtained for the derivative is $\frac{d\left(\sin x\right)}{dx}=\frac{\pi }{180}\cos \left(x\right)$ .

Explanation of Solution

Given:

Go back to the derivation of the formula for the derivative of $\sin x$ in the text, and carry out the steps of the derivation using degree-mode limits.

Considering the derivative: $\frac{d\left(\sin x\right)}{dx}$

Using the formula: $f^{\prime }\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$ here $f^{\prime }\left(x\right)$ is the first derivative of $f\left(x\right)$ and $\underset{h\to 0}{\lim }\frac{\left(\sin \left(h\right)\right)}{h}=1$ .

  $\begin{array}{c}\frac{d\left(\sin x\right)}{dx}=\underset{h\to 0}{\lim }\frac{\sin \left(x+h\right)-\sin \left(x\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{\sin \left(x\right)\cos \left(h\right)+\sin \left(h\right)\cos \left(x\right)-\sin \left(x\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{\sin \left(x\right)\left(\cos \left(h\right)-1\right)+\sin \left(h\right)\cos \left(x\right)}{h}\end{array}$

Apply limit to all the function:

  $\begin{array}{l}=\left(\underset{h\to 0}{\lim }\sin \left(x\right)\right)\left(\underset{h\to 0}{\lim }\frac{(\cos \left(h\right)-1)}{h}\right)+\left(\underset{h\to 0}{\lim }\cos \left(x\right)\right)\left(\underset{h\to 0}{\lim }\frac{\sin \left(h\right)}{h}\right)\\ =\left(\underset{h\to 0}{\lim }\sin \left(x\right)\right)\left(0\right)+\left(\underset{h\to 0}{\lim }\cos \left(x\right)\right)\left(1\right)\\ =\frac{\pi }{180}\cos \left(x\right)\end{array}$

Hence, $\frac{d\left(\sin x\right)}{dx}=\frac{\pi }{180}\cos \left(x\right)$

d.

To determine

To Derive: The formula for the derivative of $\cos x$ using degree mode limits.

Answer to Problem 50E

The formula for the derivative of $\cos x$ using degree mode limit is

  $\frac{d\left(\cos x\right)}{dx}=-\frac{\pi }{180}\sin x$ .

Explanation of Solution

Given:

Derive the formula for the derivative of $\cos x$ using degree mode limits.

Given that,

  $f\left(x\right)=\cos x$

Using the formula: $f^{\prime }\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$

Here $f^{\prime }\left(x\right)$ is the first derivative of $f\left(x\right)$ and $\underset{h\to 0}{\lim }\frac{\left(\sin \left(h\right)\right)}{h}=1$

Substituting the values into the formula:

  $\begin{array}{c}{f}^{\prime }\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{\cos \left(x+h\right)-\cos \left(x\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{\cos \left(x\right)\cos \left(h\right)-\sin \left(x\right)\sin \left(h\right)-\cos \left(x\right)}{h}\end{array}$

Apply limit to all the function:

  $\begin{array}{c}=\underset{h\to 0}{\lim }\frac{\cos \left(x\right)\left(\cos \left(h\right)-1\right)-\sin \left(x\right)\sin \left(h\right)}{h}\\ =\cos \left(x\right)\underset{h\to 0}{\lim }\frac{\left(\cos \left(h\right)-1\right)}{h}-\sin \left(x\right)\underset{h\to 0}{\lim }\frac{\left(\sin \left(h\right)\right)}{h}\\ =\cos \left(x\right)\left(0\right)-\frac{\pi }{180}\sin \left(x\right)\\ =-\frac{\pi }{180}\sin \left(x\right)\end{array}$

Hence, $\frac{d\left(\cos x\right)}{dx}=-\frac{\pi }{180}\sin x$ .

e.

To determine

To Derive: The second and third degree-mode derivatives of $\sin x$ and $\cos x$ .

Explain the disadvantages of the degree-mode formulas become even more apparent

when taking derivatives of higher order.

Answer to Problem 50E

The second order derivative of $\left(\cos x\right)$ is: $-{\left(\frac{\pi }{180}\right)}^2\left(\cos x\right)$

The third order derivative of $\left(\cos x\right)$ is: ${\left(\frac{\pi }{180}\right)}^3\left(\sin x\right)$

The second order derivative of $\left(\sin x\right)$ is: $-{\left(\frac{\pi }{180}\right)}^2\left(\sin x\right)$

The third order derivative of $\left(\sin x\right)$ is: $-{\left(\frac{\pi }{180}\right)}^3\left(\cos x\right)$

Explanation of Solution

Given:

Derive the second and third degree-mode derivatives of $\sin x$ and $\cos x$ .

Given that the second and third order derivative of $\left(\cos x\right)$ ,

The second order derivative of $\left(\cos x\right)$ is:

  $\begin{array}{c}\frac{d^2\left(\cos x\right)}{d{x}^2}=\frac{d\left(-\frac{\pi }{180}\sin x\right)}{dx}\\ =-\frac{\pi }{180}\frac{d\left(\sin x\right)}{dx}\\ =-{\left(\frac{\pi }{180}\right)}^2\left(\cos x\right)\end{array}$

The third order derivative of $\left(\cos x\right)$ is:

  $\begin{array}{c}\frac{d^3\left(\cos x\right)}{d{x}^3}=\frac{d\left(-{\left(\frac{\pi }{180}\right)}^2\left(\cos x\right)\right)}{dx}\\ ={\left(\frac{\pi }{180}\right)}^3\left(\sin x\right)\end{array}$

Likewise, the second order derivative of $\left(\sin x\right)$ is:

  $\begin{array}{c}\frac{d^2\left(\sin x\right)}{d{x}^2}=\frac{d\left(\frac{\pi }{180}\cos x\right)}{dx}\\ =\frac{\pi }{180}\frac{d\left(\cos x\right)}{dx}\\ =-{\left(\frac{\pi }{180}\right)}^2\left(\sin x\right)\end{array}$

The third order derivative of $\left(\sin x\right)$ is:

  $\begin{array}{c}\frac{d^3\left(\sin x\right)}{d{x}^3}=\frac{d\left(-{\left(\frac{\pi }{180}\right)}^2\left(\sin x\right)\right)}{dx}\\ =-{\left(\frac{\pi }{180}\right)}^3\left(\cos x\right)\end{array}$