(vii) Assume, $f(x) = 2\tan x - 7\sec x$. Using the first principle we get, ${f^\prime }(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h}$ ${f^\prime }(x) = \mathop {\lim...
(vi) 
(v) Assume, $f(x) = 3\cot x + 5\operatorname{cosec} x$. Upon differentiating with respect to $x$ to get, ${{\text{f}}^\prime }({\text{x}}) = 3{(\cot x)^\prime } + 5{(\operatorname{cosec}...
(iv) 
(iii) Assume, $f(x) = 5\sec x + 4\cos x$. Upon differentiating with respect to $x$ to get, ${{\text{f}}^\prime }({\text{x}}) = \frac{{\text{d}}}{{{\text{dx}}}}(5\sec {\text{x}} + 4\cos {\text{x}})$...
at 
.Assume, $f(x) = 99x$, By the first principle we have, ${{\text{f}}^\prime }({\text{x}}) = \mathop {\lim }\limits_{{\text{h}} \to 0} \frac{{{\text{f}}({\text{x}} + {\text{h}}) -...
(ii) 
.(i) Assume, $f{\text{ }}\left( x \right){\text{ }} = {\text{ }}{x^3}\;--{\text{ }}27$. Using first principle we have, ${{\text{f}}^\prime }({\text{x}}) = \mathop {\lim }\limits_{{\text{h}} \to 0}...
(iv) 
(iii) Assume, $\;f\left( x \right) = \frac{1}{{{x^2}}}$. Using first principle we have, ${f^\prime }(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h}$ $f'(x) = \mathop {\lim...
.Prove that 
.We are given, $f(x) = \frac{{{x^{100}}}}{{100}} + \frac{{{x^{99}}}}{{99}} + \ldots \frac{{{x^2}}}{2} + x + 1$ Differentiating with respect to $x$ to get, $\frac{d}{{dx}}f(x) = \frac{d}{{dx}}\left[...
for some fixed real number 
.We are given, ${x^n} + a{x^{n - 1}} + {a^2}{x^{n - 2}} + \ldots + {a^{n - 1}}x + {a^n}$. Differentiating with respect to $x$ to get, ${f^\prime }(x) = \frac{d}{{dx}}\left( {{x^n} + a{x^{n - 1}} +...
(ii) 
.(i) Assume, ${\text{f(x) = (x - a)(x - b)}}$ ${\text{f}}({\text{x}}) = {{\text{x}}^2} - ({\text{a}} + {\text{b}}){\text{x}} + {\text{ab}}$ Differentiating with respect to $x$ to get,...
(iii) Assume, $f(x) = \frac{{(x - a)}}{{(x - b)}}$. Upon differentiating both sides and using quotient rule to get, ${f^\prime }(x) = \frac{d}{{dx}}\left( {\frac{{x - a}}{{x - b}}} \right)$...
for some constant .Assume, $f(x) = \frac{{{x^n} - {a^n}}}{{x - a}}$. Upon differentiating both sides and using quotient rule to get, ${f^\prime }(x) = \frac{d}{{dx}}\left( {\frac{{{x^n} - {a^n}}}{{x - a}}} \right)$...
(ii) 
.(i) Assume, $f(x) = \frac{{2x - 3}}{4}$. Differentiating with respect to $x$ to get, ${f^\prime }(x) = \frac{d}{{dx}}\left( {2x - \frac{3}{4}} \right)$ ${f^\prime }(x) = 2\frac{d}{{dx}}(x) -...
(iv) 
(iii) Assume, $f(x) = {x^{ - 3}}(5 + 3x)$. Upon differentiating with respect to $x$ and applying Leibnitz product rule to get, ${f^\prime }(x) = {x^{ - 3}}\frac{d}{{dx}}(5 + 3x) + (5 +...
(vi) 
(v) Assume, $f(x) = {x^{ - 4}}\left( {3 - 4{x^{ - 5}}} \right)$. Upon differentiating with respect to $x$ and applying Leibnitz product rule to get, ${f^\prime }(x) = {x^{ - 4}}\frac{d}{{dx}}\left(...
(ii) 
.(i) Assume, $f(x) = \sin x\cos x$. Using the first principle we get, ${f^\prime }(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h}$ $f'(x) = \mathop {\lim }\limits_{h \to 0}...
from first principle.Assume, $f(x) = \cos x$. Then, $f(x + h) = \cos (x + h)$. Using first principle we get, ${{\text{f}}^\prime }({\text{x}}) = \mathop {\lim }\limits_{h \to 0} \frac{{{\text{f}}({\text{x}} +...
at 
.Suppose $f(x) = x$. By the first principle we have, ${{\text{f}}^\prime }({\text{x}}) = \mathop {\lim }\limits_{{\text{h}} \to 0} \frac{{{\text{f}}({\text{x}} + {\text{h}}) -...
at 
.Suppose $f(x) = {x^2} - 2$. By the first principle we have, ${{\text{f}}^\prime }({\text{x}}) = \mathop {\lim }\limits_{{\text{h}} \to 0} \frac{{{\text{f}}({\text{x}} + {\text{h}}) -...