Q. 19 - Calculus

Question

Consider the function 

$F\left(x\right)=\left\{\begin{array}{l} -cot x, if x<0\\ -cot x+100, if x>0\end{array}\right.$

Show that the derivative of this function is the function $f\left(x\right)=cs{c}^{2}x$. Compare the graphs of $F\left(x\right)$ and $-cotx$, and discuss how this exercise relates to the fourth part of Theorem 4.18.

Step-by-Step Solution

Answer

Proved that the derivative of this function is the function $f\left(x\right)=cs{c}^{2}x$

Step 1. Given information

The given function $F\left(x\right)=\left\{\begin{array}{l} -cot x, if x<0\\ -cot x+100, if x>0\end{array}\right.$

Step 2. Draw graph for - c o t &#160; x for all x and - c o t &#160; x + 100 for x &#62; 0

The graph of $-cotx and -cotx+100$

Step 3. Finding the differentiation of c o t &#160; x

First consider the differentiation of $cot x$

$$\begin{gathered} =\frac{d cotx}{dx} \\ =\frac{d}{dx}\left[\frac{\cos x}{\sin x}\right] \\ =\frac{\sin x(-\sin x)-\cos x\left(\cos x\right)}{{\left(\sin x\right)}^2} \\ =\frac{-{\sin }^{2}x-{\cos }^{2}x}{{\sin }^{2}x} \\ =-\frac{1}{{\sin }^{2}x} \\ =-cs{c}^{2}x \end{gathered}$$

Step 4. The derivative of the function F ( x ) can be given as shown below

$F\text{'}\left(x\right)=\left\{\begin{array}{l}-\frac{dcotx}{dx} if;x<0\\ -\frac{dcotx}{dx}+\frac{d}{dx}\left(100\right) if;x>0\end{array}\right.$

$$\begin{gathered} F\text{'}\left(x\right)=\left\{\begin{array}{l}-(-cs{c}^{2}x) if;x<0\\ -\left(cs{c}^{2}x\right)+0 if;x>0\end{array}\right. \\ F\text{'}\left(x\right)=\left\{\begin{array}{l}cs{c}^{2}x if;x<0\\ cs{c}^{2}x if;x>0\end{array}\right. \\ F\text{'}\left(x\right)=cs{c}^{2}x \end{gathered}$$

Now,

$\frac{d(-\mathrm{co}tx)}{dx}=cs{c}^{2}x$

Thus gradient of the function $F\left(x\right)$ and gradient of the function $-cotx$ are same and it is $cs{c}^{2}x$