Tag Archives: Differentiation
f(x) + f(y) + f(x)f(y) = 1For all values of $x,y \in R$f'(x) = ?
Putting y = x we have,
$2f(x) + {\left[ {f(x)} \right]^2} = 1$
$2f(x) + {\left[ {f(x)} \right]^2} + 1 = 1 + 1$
${[f(x) + 1]^2} = 2$
$f(x) = – 1 \pm \sqrt 2 $
Since f(x) is a constant function, $f'(x) = 0$.
f'(0) = ? for an even function f that is differentiable at 0
Since f is differentiable at 0 we have,
$\mathop {\lim }\limits_{h \to 0} \frac{{f(0 + h) – f(0)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{f(0 – h) – f(0)}}{{ – h}}$
$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{f(h) – f(0)}}{h} = – \mathop {\lim }\limits_{h \to 0} \frac{{f( – h) – f(0)}}{h}$
Since f is an even function, f(-h) = f(h) Continue reading f'(0) = ? for an even function f that is differentiable at 0
Let $I = \int\limits_0^{\pi /2} {\frac{{dx}}{{1 + {{\tan }^{\sqrt {\tan \alpha } }}x}}}$$\frac{{dI}}{{d\alpha }} = ?$
We have, $I = \int\limits_0^{\pi /2} {\frac{{dx}}{{1 + {{\tan }^{\sqrt {\tan \alpha } }}(\pi /2 – x)}}} $
$ \Rightarrow I = \int\limits_0^{\pi /2} {\frac{{dx}}{{1 + {{\cot }^{\sqrt {\tan \alpha } }}x}}} = \int\limits_0^{\pi /2} {\frac{{{{\sin }^{\sqrt {\tan \alpha } }}x.dx}}{{{{\sin }^{\sqrt {\tan \alpha } }}x + {{\cos }^{\sqrt {\tan \alpha } }}x}}} $ ……..(A)
Also, $I = \int\limits_0^{\pi /2} {\frac{{dx}}{{1 + {{\tan }^{\sqrt {\tan \alpha } }}x}} = } \int\limits_0^{\pi /2} {\frac{{{{\cos }^{\sqrt {\tan \alpha } }}x.dx}}{{{{\cos }^{\sqrt {\tan \alpha } }}x + {{\sin }^{\sqrt {\tan \alpha } }}x}}}$ ……..(B)
(A) + (B) gives, $2I = \int\limits_0^{\pi /2} {dx} = \frac{\pi }{2}$
$ \Rightarrow I = \frac{\pi }{4}$
$\therefore \frac{{dI}}{{d\alpha }} = 0$
Differentiation Quiz
If f(x) = x + tan x and $$g = f^{-1}$$, then g'(x) is
If f(x) = x + tan x and g = f-1, then g'(x) is equal to
A) 
B) 
C) 
D) 
Continue reading If f(x) = x + tan x and $$g = f^{-1}$$, then g'(x) is