(vii) Assume, 
Using the first principle we get,
![{f^\prime }(x) = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}[2\tan (x + h) - 7\sec (x + h) - 2\tan x + 7\sec x]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-77d115e52324debf0ddb22cee49ca15c_l3.png "Rendered by QuickLaTeX.com")
We get,
![{f^\prime }(x) = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}[2\{ \tan (x + h) - \tan x\} - 7\{ \sec (x + h) - \sec x\} ]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a485dcd07268b2a020596dd7eb3c8c2c_l3.png "Rendered by QuickLaTeX.com")
![{f^\prime }(x) = 2\mathop {\lim }\limits_{h \to 0} \frac{1}{h}[\tan (x + h) - \tan x] - 7\mathop {\lim }\limits_{h \to 0} \frac{1}{h}[\sec (x + h) - \sec x]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c98cb2482ebf22f45bef89dbbaff3c45_l3.png "Rendered by QuickLaTeX.com")
Further simplifying to get,
![f'(x) = 2\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{\sin (x + h)}}{{\cos (x + h)}} - \frac{{\sin x}}{{\cos x}}} \right] - 7\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{1}{{\cos (x + h)}} - \frac{1}{{\cos x}}} \right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-45bc624f43ff4ab12de5febddd9e8922_l3.png "Rendered by QuickLaTeX.com")
![f'(x) = 2\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{\sin (x + h)\cos x - \sin x\cos (x + h)}}{{\cos x\cos (x + h)}}} \right] - 7\mathop {\lim }\limits_{k \to 0} \frac{1}{h}\left[ {\frac{{\cos x - \cos (x + h)}}{{\cos x\cos (x + h)}}} \right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e18d41a9351a4b88f010991c8a3a6952_l3.png "Rendered by QuickLaTeX.com")
![f'(x) = 2\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{\sin (x + h - x)}}{{\cos x\cos (x + h)}}} \right] - 7\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{ - 2\sin \left( {\frac{{x + x + h}}{2}} \right)\sin \left( {\frac{{x - x - h}}{2}} \right)}}{{\cos x\cos (x + h)}}} \right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-eaa6df138f592b56772d403f3fed5489_l3.png "Rendered by QuickLaTeX.com")
Upon simplifying we get,
![f'(x) = 2\mathop {\lim }\limits_{h \to 0} \left[ {\left( {\frac{{\sin h}}{h}} \right)\frac{1}{{\cos x\cos (x + h)}}} \right] - 7\mathop {\lim }\limits_{b \to 0} \frac{1}{h}\left[ {\frac{{ - 2\sin \left( {\frac{{2x + h}}{2}} \right)\sin \left( { - \frac{h}{2}} \right)}}{{\cos x\cos (x + h)}}} \right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-9337a139501b17ef67a517d7da6f98d6_l3.png "Rendered by QuickLaTeX.com")
Therefore, 
Find the derivative of the following functions: (vii) 
(vii) Assume,
Using the first principle we get,
We get,
Further simplifying to get,
Upon simplifying we get,
Therefore,
