Question

If$f(x) = \operatorname{Cos} ecx + Cotx$, then the value of $f'\left( {\dfrac{\pi }{4}} \right)$ is

Answer 1

Hint: The differentiation of the function is to be performed and the value should be substituted. Use relevant formulae of differentiation of the given function and once you find differentiation of the function substitute x=pi/4 to get the required value.


Complete step-by-step answer:
The given function is
 $f(x) = \operatorname{Cos} ecx + Cotx \cdots \left( 1 \right)$
The value of the derivative is to be found at $\dfrac{\pi }{4}$
Differentiation operations should be performed on the given function.
The function given is the addition of two functions and .The two functions should be differentiable at the given point.
Also the value of the derivative of these functions should be remembered.
The individual derivative of the functions is
$y = \cos ecx$ is
 $\dfrac{{dy}}{{dx}} = - \cos ecx.\cot x$
$y = \cot x$
$\dfrac{{dy}}{{dx}} = - \cos e{c^2}x$
Now, perform the differentiation operation of equation (1) concerning to x,
\[f'\left( x \right) = - \cos ecx.\cot x - \cos e{c^2}x\]
Taking common from the two terms,
\[f'\left( x \right) = - \cos ecx\left( {\cot x + \cos ecx} \right) \cdots \left( 2 \right)\]
Substitute the $x = \dfrac{\pi }{4}$ in equation (2),
$f'\left( {\dfrac{\pi }{4}} \right) = - \cos ec\left( {\dfrac{\pi }{4}} \right)\left( {\cot \left( {\dfrac{\pi }{4}} \right) + \cos ec\left( {\dfrac{\pi }{4}} \right)} \right) \cdots \left( 3 \right)$
The value of $\cos ec\left( {\dfrac{\pi }{4}} \right) = \sqrt 2 $ and $\cot \left( {\dfrac{\pi }{4}} \right) = 1$ .
Substituting the value of $\cos ec\left( {\dfrac{\pi }{4}} \right)$ and $\cot \left( {\dfrac{\pi }{4}} \right)$ in equation (3)
$f'\left( {\dfrac{\pi }{4}} \right) = - \sqrt 2 \left( {1 + \sqrt 2 } \right)$

Note: The important point before performing the differentiation operation is to check whether a function is differentiable or not.
Also the formula for differentiation of various trigonometric identities should be remembered.
The differentiation of some important trigonometric functions is,
$\dfrac{{d\left( {\sin x} \right)}}{{dx}} = \cos x$
$\dfrac{{d\left( {\cos x} \right)}}{{dx}} = - \sin x$
$\dfrac{{d\left( {\tan x} \right)}}{{dx}} = {\sec ^2}x$
$\dfrac{{d\left( {\cos ecx} \right)}}{{dx}} = - \cos ecx.\cot x$
$\dfrac{{d\left( {\cot x} \right)}}{{dx}} = - \cos e{c^2}x$