Question

What is the derivative of $\cos ec\left( x \right)$ ?

Answer 1

Hint : The given question wants us to evaluate the derivative of a given trigonometric function using the first principle of derivative. The first principle of differentiation involves the concepts of limits, derivatives, continuity and differentiability. It helps us to determine the derivative of the function using concepts of limits.

Complete step by step solution:
Given the function $\cos ec\left( x \right)$, we have to calculate the derivative of the function using the first principle of derivatives.
Using formula of first principle of derivatives, we know that derivative of a function is calculated as:
$f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}$
So, derivative of the given function $f(x) = \cos ec\left( x \right)$ ,
$\Rightarrow$ $f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos ec\left( {x + h} \right) - \cos ec\left( x \right)}}{h}$
We know that $\sin x$ and $\cos ec\left( x \right)$ are reciprocals of each other. Hence, we get,
$\Rightarrow$ $f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\dfrac{1}{{\sin \left( {x + h} \right)}} - \dfrac{1}{{\sin \left( x \right)}}}}{h}$
Taking LCM, we get,
$\Rightarrow$ $f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\dfrac{{\sin \left( x \right) - \sin \left( {x + h} \right)}}{{\sin \left( {x + h} \right)\sin \left( x \right)}}}}{h}$
Expanding the compound angle formula of sine, we get,
$\Rightarrow$ $f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\dfrac{{\sin \left( x \right) - \left[ {\sin \left( x \right)\cos \left( h \right) + \sin \left( h \right)\cos \left( x \right)} \right]}}{{\sin \left( {x + h} \right)\sin \left( x \right)}}}}{h}$
$\Rightarrow$ f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left\\{ {\sin \left( x \right) - \sin \left( x \right)\cos \left( h \right)} \right\\} - \sin \left( h \right)\cos \left( x \right)}}{{h\,\sin \left( {x + h} \right)\sin \left( x \right)}}
Separating the denominators, we get,
$\Rightarrow$ f'(x) = \mathop {\lim }\limits_{h \to 0} \left[ {\dfrac{{\sin \left( x \right)\left\\{ {1 - \cos \left( h \right)} \right\\}}}{{h\,\sin \left( {x + h} \right)\sin \left( x \right)}} - \dfrac{{\sin \left( h \right)\cos \left( x \right)}}{{h\,\sin \left( {x + h} \right)\sin \left( x \right)}}} \right]
Cancelling the common factors in numerator and denominator and separating two limits, we get,
$\Rightarrow$ f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left\\{ {1 - \cos \left( h \right)} \right\\}}}{{h\,\sin \left( {x + h} \right)}} - \mathop {\lim }\limits_{h \to 0} \dfrac{{\sin \left( h \right)\cos \left( x \right)}}{{h\,\sin \left( {x + h} \right)\sin \left( x \right)}}
Now, substituting the value of $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x}$ as $1$, we get,
$\Rightarrow$ f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left\\{ {1 - \cos \left( h \right)} \right\\}}}{{h\,\sin \left( {x + h} \right)}} - \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos \left( x \right)}}{{\sin \left( {x + h} \right)\sin \left( x \right)}}
We know that $1 - \cos x = 2{\sin ^2}\left( {\dfrac{x}{2}} \right)$ . So, we get,
$\Rightarrow$ $f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{2{{\sin }^2}\left( {\dfrac{h}{2}} \right)}}{{h\,\sin \left( {x + h} \right)}} - \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos \left( x \right)}}{{\sin \left( {x + h} \right)\sin \left( x \right)}}$
Simplifying further, we get,
$\Rightarrow$ $f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\sin \left( {\dfrac{h}{2}} \right)\sin \left( {\dfrac{h}{2}} \right)}}{{\dfrac{h}{2}\,\sin \left( {x + h} \right)}} - \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos \left( x \right)}}{{\sin \left( {x + h} \right)\sin \left( x \right)}}$
$\Rightarrow$ $f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\sin \left( {\dfrac{h}{2}} \right)}}{{\,\sin \left( {x + h} \right)}} - \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos \left( x \right)}}{{\sin \left( {x + h} \right)\sin \left( x \right)}}$
Now, putting in the value of h in the limit, we get,
$\Rightarrow$ $f'(x) = \dfrac{{\sin \left( {\dfrac{0}{2}} \right)}}{{\,\sin \left( {x + 0} \right)}} - \dfrac{{\cos \left( x \right)}}{{\sin \left( {x + 0} \right)\sin \left( x \right)}}$
$\Rightarrow$ $f'(x) = 0 - \dfrac{{\cos \left( x \right)}}{{\sin \left( x \right)\sin \left( x \right)}}$
$\Rightarrow$ $f'(x) = - \cot \left( x \right)\cos ec\left( x \right)$
Therefore, we get the derivative of $\cos ec\left( x \right)$ as $- \cot \left( x \right)\cos ec\left( x \right)$ using the first principle of differentiation.

Note : There are various methods of finding the derivatives of the given functions. The calculation of derivative using the first principle of derivative is the most basic and core method to find the derivative of a specific function. Another method of finding derivatives of a function can be simply differentiating that particular function with respect to the variable or unknown using some basic differentiation rules like addition rule, subtraction rule, product rule, and quotient rule. This method is comparatively easier as it does not involve concepts limits and continuity.