(i) Assume, 
Using the first principle we get,
![f'(x) = \mathop {\lim }\limits_{h \to 0} \frac{1}{{2h}}[2\sin (x + h)\cos (x + h) - 2\sin x\cos x]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c07e32beb8ed195a2f6036e2d8cb0631_l3.png "Rendered by QuickLaTeX.com")
![f'(x) = \mathop {\lim }\limits_{h \to 0} \frac{1}{{2h}}[\sin 2(x + h) - \sin 2x]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ec5f933f66dd52331e33e0b15c560c93_l3.png "Rendered by QuickLaTeX.com")
![f'(x) = \mathop {\lim }\limits_{h \to 0} \frac{1}{{2h}}\left[ {2\cos \frac{{2x + 2h + 2x}}{2} \cdot \sin \frac{{2x + 2h - 2x}}{2}} \right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-665d24cd3f0fd7659355bee32f7d54ba_l3.png "Rendered by QuickLaTeX.com")
![f'(x) = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\cos \frac{{4x + 2h}}{2}\sin \frac{{2h}}{2}} \right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-60b4d2e7a6f647094587827ca27eca5e_l3.png "Rendered by QuickLaTeX.com")
![f'(x) = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}[\cos (2x + h)\sin h]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-45dc9e59120434925f2b394ff4a81f19_l3.png "Rendered by QuickLaTeX.com")
(ii) Assume, 
Upon differentiating with respect to 
Therefore, 
Find the derivative of the following functions: (i) 
(i) Assume,
Using the first principle we get,
(ii) Assume,
Upon differentiating with respect to
Therefore,
