Question

Find the derivative of the following w. r. t. x. at the point indicated against them by using method of first principle:

cos x at x = ${\displaystyle \frac{5\pi }{4}}$

Answer 1

Let f(x) = cos x

∴ ${\displaystyle \text{f}\left(\frac{5\pi }{4}\right)=\cos \left(\frac{5\pi }{4}\right)}$

${\displaystyle \text{f}(\frac{5\pi }{4}+\text{h})=\cos (\frac{5\pi }{4}+\text{h})}$

∴ ${\displaystyle \text{f}(\frac{5\pi }{4}+\text{h})-\text{f}\left(\frac{5\pi }{4}\right)}$

= ${\displaystyle \cos (\frac{5\pi }{4}+\text{h})-\cos \left(\frac{5\pi }{4}\right)}$

= ${\displaystyle -2\sin \left[\frac{\frac{5\pi }{4}+\text{h}+\frac{5\pi }{4}}{2}\right]\sin \left[\frac{\frac{5\pi }{4}+\text{h}-\frac{5\pi }{4}}{2}\right]}$

= ${\displaystyle -2\sin \left[\frac{\frac{5\pi }{2}+\text{h}}{2}\right]\sin \left(\frac{\text{h}}{2}\right)}$

By definition,

${\displaystyle \text{f'}\left(\frac{5\pi }{4}\right)= \underset{\text{h}\to 0}{lim}\frac{\text{f}(\frac{5\pi }{4}+\text{h})-\text{f}\left(\frac{5\pi }{4}\right)}{\text{h}}}$

= ${\displaystyle \underset{\text{h}\to 0}{lim}\frac{-2\sin \left[\frac{\frac{5\pi }{2}+\text{h}}{2}\right]\sin \left(\frac{\text{h}}{2}\right)}{\text{h}}}$

= ${\displaystyle \underset{\text{h}\to 0}{lim}[-2\sin \left[\frac{\frac{5\pi }{2}+\text{h}}{2}\right]\frac{\sin \left(\frac{\text{h}}{2}\right)}{\left(\frac{\text{h}}{2}\right)}]\times \frac{1}{2}}$

= ${\displaystyle -\left[\underset{\text{h}\to 0}{lim}\sin \left[\frac{\frac{5\pi }{2}+\text{h}}{2}\right]\right]\times \left[\underset{\text{h}\to 0}{lim}\frac{\sin \left(\frac{\text{h}}{2}\right)}{\frac{\text{h}}{2}}\right]}$

= ${\displaystyle -\sin \left[\frac{\frac{5\pi }{2}+0}{2}\right]\times 1 ...[\because \text{h}\to 0,\frac{\text{h}}{2}\to 0 \text{and}\underset{\theta \to 0}{lim}\frac{\sin \theta }{\theta }=1]}$

= ${\displaystyle -\sin \left(\frac{5\pi }{4}\right)}$

= ${\displaystyle -\sin (\pi +\frac{\pi }{4})}$

= ${\displaystyle -(-\sin \frac{\pi }{4})}$

= ${\displaystyle \frac{1}{\sqrt{2}}}$