Question

Find the derivative of $\cos e{c^2}x$, by using the first principal of derivatives?

Answer 1

Hint: In this particular type of question use the concept that to find the differentiation by first principle derivatives the formula is given as \[\dfrac{d}{{dx}}f\left( x \right) = f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}\] and later on use the concept that $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1$ so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Given equation:
$\cos e{c^2}x$
Let,
$ \Rightarrow f\left( x \right) = \cos e{c^2}x$............... (1)
Now we have to find its derivation using first principal of derivatives,
So according to first principal the derivative of function f(x) is given as,
\[ \Rightarrow \dfrac{d}{{dx}}f\left( x \right) = f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}\]............... (2)
So from equation (1) value of f (x + h) = $\cos e{c^2}\left( {x + h} \right)$............... (3)
Now substitute the value from equation (1) and (3) in equation (2) we have,
\[ \Rightarrow \dfrac{d}{{dx}}f\left( x \right) = f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos e{c^2}\left( {x + h} \right) - \cos e{c^2}x}}{h}\]
Now as we know that cosec x = (1/sin x) so use this property in the above equation we have,
\[ \Rightarrow \dfrac{d}{{dx}}f\left( x \right) = f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\dfrac{1}{{{{\sin }^2}\left( {x + h} \right)}} - \dfrac{1}{{{{\sin }^2}x}}}}{h}\]
Now simplify this we have,
\[ \Rightarrow \dfrac{d}{{dx}}f\left( x \right) = f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{{{\sin }^2}x - {{\sin }^2}\left( {x + h} \right)}}{{h{{\sin }^2}\left( {x + h} \right){{\sin }^2}x}}\]
Now use the property that $\left( {{a^2} - {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right)$ in the above equation we have,
\[ \Rightarrow \dfrac{d}{{dx}}f\left( x \right) = f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {\sin x - \sin \left( {x + h} \right)} \right)\left( {\sin x + \sin \left( {x + h} \right)} \right)}}{{h{{\sin }^2}\left( {x + h} \right){{\sin }^2}x}}\]
Now as we know that $\sin C - \sin D = 2\sin \dfrac{{C - D}}{2}\cos \dfrac{{C + D}}{2}$ so use this property in the above equation we have,
\[ \Rightarrow \dfrac{d}{{dx}}f\left( x \right) = f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {2\sin \dfrac{{x - x - h}}{2}\cos \dfrac{{x + x + h}}{2}} \right)\left( {\sin x + \sin \left( {x + h} \right)} \right)}}{{h{{\sin }^2}\left( {x + h} \right){{\sin }^2}x}}\]
Now simplify it we have,

\[ \Rightarrow \dfrac{d}{{dx}}f\left( x \right) = f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {2\sin \dfrac{{ - h}}{2}\cos \dfrac{{2x + h}}{2}} \right)\left( {\sin x + \sin \left( {x + h} \right)} \right)}}{{h{{\sin }^2}\left( {x + h} \right){{\sin }^2}x}}\]
Now as we know that sin (-x) = -sin x so we have,
\[ \Rightarrow \dfrac{d}{{dx}}f\left( x \right) = f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{ - \left( {2\sin \dfrac{h}{2}\cos \dfrac{{2x + h}}{2}} \right)\left( {\sin x + \sin \left( {x + h} \right)} \right)}}{{h{{\sin }^2}\left( {x + h} \right){{\sin }^2}x}}\]
Now multiply and divide by 2 in denominator h term so we have,
\[ \Rightarrow \dfrac{d}{{dx}}f\left( x \right) = f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{ - \left( {2\sin \dfrac{h}{2}\cos \dfrac{{2x + h}}{2}} \right)\left( {\sin x + \sin \left( {x + h} \right)} \right)}}{{2\left( {\dfrac{h}{2}} \right){{\sin }^2}\left( {x + h} \right){{\sin }^2}x}}\]
\[ \Rightarrow \dfrac{d}{{dx}}f\left( x \right) = f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} - \left( {\dfrac{{\sin \dfrac{h}{2}}}{{\dfrac{h}{2}}}} \right)\dfrac{{\left( {\cos \dfrac{{2x + h}}{2}} \right)\left( {\sin x + \sin \left( {x + h} \right)} \right)}}{{{{\sin }^2}\left( {x + h} \right){{\sin }^2}x}}\]
Now as we know that \[\mathop {\lim }\limits_{h \to 0} \left( {\dfrac{{\sin \dfrac{h}{2}}}{{\dfrac{h}{2}}}} \right) = 1\], so use this property in the above equation we have,
\[ \Rightarrow \dfrac{d}{{dx}}f\left( x \right) = f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} - \dfrac{{\left( {\cos \dfrac{{2x + h}}{2}} \right)\left( {\sin x + \sin \left( {x + h} \right)} \right)}}{{{{\sin }^2}\left( {x + h} \right){{\sin }^2}x}}\]
Now apply the limit i.e. substitute h = 0 we have,
\[ \Rightarrow \dfrac{d}{{dx}}f\left( x \right) = f'\left( x \right) = - \dfrac{{\left( {\cos x} \right)\left( {\sin x + \sin x} \right)}}{{{{\sin }^2}x{{\sin }^2}x}}\]
Now simplify this we have,
\[ \Rightarrow \dfrac{d}{{dx}}f\left( x \right) = f'\left( x \right) = - \dfrac{{2\left( {\cos x} \right)\left( {\sin x} \right)}}{{{{\sin }^2}x{{\sin }^2}x}}\]
\[ \Rightarrow \dfrac{d}{{dx}}f\left( x \right) = f'\left( x \right) = - \dfrac{{2\left( {\cos x} \right)}}{{{{\sin }^2}x\sin x}}\]
\[ \Rightarrow \dfrac{d}{{dx}}f\left( x \right) = f'\left( x \right) = - 2\cos e{c^2}x\cot x\], $\left[ {\because \dfrac{{\cos x}}{{\sin x}} = \cot x,\dfrac{1}{{{{\sin }^2}x}} = \cos e{c^2}x} \right]$
So this is the required differentiation of \[\cos e{c^2}x\] using the first principle of derivatives.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the formula of first derivative which is stated above, also remember that $\left( {{a^2} - {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right)$ and the basic trigonometric identity such as $\sin C - \sin D = 2\sin \dfrac{{C - D}}{2}\cos \dfrac{{C + D}}{2}$ to get the required solution as above.