Chapter 3.5, Problem 24E

To determine

To use:The derivative to prove that $(\frac{d}{dx})\cos x=-\sin x$ .

Answer to Problem 24E

  $\begin{array}{l}\frac{d}{dx}\cos x=\underset{h\to 0}{{\displaystyle \lim }}\frac{\cos (x+h)-\cos x}{h}\\ \frac{d}{dx}\cos x=\underset{h\to 0}{{\displaystyle \lim }}\frac{\cos x\cosh -\sin x\sinh -\cos x}{h}\\ \frac{d}{dx}\cos x=\underset{h\to 0}{{\displaystyle \lim }}(\frac{\cos x\cosh -\cos x}{h}-\frac{\sin x\sinh }{h})\\ \frac{d}{dx}\cos x=\underset{h\to 0}{{\displaystyle \lim }}(\cos x.\frac{\cosh -1}{h}-\sin x.\frac{\sinh }{h})\\ \frac{d}{dx}\cos x=\underset{h\to 0}{{\displaystyle \lim }}(\cos x.\frac{\cosh -1}{h})-\underset{h\to 0}{{\displaystyle \lim }}(\sin x.\frac{\sinh }{h})\\ \frac{d}{dx}\cos x=\cos x\underset{h\to 0}{{\displaystyle \lim }}\frac{\cosh -1}{h}-\sin x.\underset{h\to 0}{{\displaystyle \lim }}\frac{\sinh }{h}\\ \frac{d}{dx}\cos x=\cos x.0-\sin x.1=-\sin x\end{array}$

Explanation of Solution

Given information:Use the definition of the derivative to prove that $(\frac{d}{dx})\cos x=-\sin x$

Calculation:Using the definition of derivative:

  $\frac{d}{dx}f(x)=f\text{'}(x)=\underset{h\to 0}{{\displaystyle \lim }}\frac{f(x+h)-f(x)}{h}$

  $\begin{array}{l}\frac{d}{dx}\cos x=\underset{h\to 0}{{\displaystyle \lim }}\frac{\cos (x+h)-\cos x}{h}\\ \frac{d}{dx}\cos x=\underset{h\to 0}{{\displaystyle \lim }}\frac{\cos x\cosh -\sin x\sinh -\cos x}{h}\\ \frac{d}{dx}\cos x=\underset{h\to 0}{{\displaystyle \lim }}(\frac{\cos x\cosh -\cos x}{h}-\frac{\sin x\sinh }{h})\\ \frac{d}{dx}\cos x=\underset{h\to 0}{{\displaystyle \lim }}(\cos x.\frac{\cosh -1}{h}-\sin x.\frac{\sinh }{h})\\ \frac{d}{dx}\cos x=\underset{h\to 0}{{\displaystyle \lim }}(\cos x.\frac{\cosh -1}{h})-\underset{h\to 0}{{\displaystyle \lim }}(\sin x.\frac{\sinh }{h})\\ \frac{d}{dx}\cos x=\cos x\underset{h\to 0}{{\displaystyle \lim }}\frac{\cosh -1}{h}-\sin x.\underset{h\to 0}{{\displaystyle \lim }}\frac{\sinh }{h}\\ \frac{d}{dx}\cos x=\cos x.0-\sin x.1=-\sin x\end{array}$

Therefore $(\frac{d}{dx})\cos x=-\sin x$