Chapter 4.1, Problem 77E

To determine

To calculate: To graph the function $y=-2x\sin x^2$ for $-2\le x\le 3$ and graph $y=\frac{\cos {\left(x+h\right)}^2-\cos x^2}{h}$ for $h=1,0.7,0.3$ . What happens as $h\to 0$

Answer to Problem 77E

As $h\to 0$, the function approaches to $-2x\sin x^2$.

Explanation of Solution

Given Information: The functions are $y=-2x\sin x^2$ and $y=\frac{\cos {\left(x+h\right)}^2-\cos x^2}{h}$

Calculation:

The graph of the above equations is,

  

The black curve represents the graph of the curve $y=-2x\sin x^2$ .

Now, if the negative values are used for h, the values will approach the function from right- hand side instead from the left-hand side.

Using the definition of the derivative of the function,

  $\begin{array}{c}\frac{d}{dx}f\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}\\ \frac{d}{dx}\cos x^2=\underset{h\to 0}{\lim }\frac{\cos {\left(x+h\right)}^2-\cos x^2}{h}\\ -2x\sin x^2=\underset{h\to 0}{\lim }\frac{\cos {\left(x+h\right)}^2-\cos x^2}{h}\end{array}$

Thus, as $h\to 0$ , the function approaches to $-2x\sin x^2$ .