Question

Differentiate $\sec x$ by using first principle.

Answer 1

Hint- According to the first principle of differentiation, If we differentiate a function f(x) so the differentiation f’(x) represents the value of slope of a tangent on a curve. We use formula of differentiation \[f'\left( x \right) = \dfrac{d}{{dx}}\left( {f\left( x \right)} \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f(x)}}{h}\].

Complete step-by-step solution -
Now, we have to differentiate $\sec x$ by using first principle.
Let $f\left( x \right) = \sec x$
First principal of differentiation,
\[f'\left( x \right) = \dfrac{d}{{dx}}\left( {f\left( x \right)} \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f(x)}}{h}\]
Now, $f\left( x \right) = \sec x$ and $f\left( {x + h} \right) = \sec \left( {x + h} \right)$
Put the value of f(x) and f(x+h)
\[
   \Rightarrow \dfrac{d}{{dx}}\left( {\sec x} \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\sec \left( {x + h} \right) - \sec x}}{h} \\
   \Rightarrow \dfrac{d}{{dx}}\left( {\sec x} \right) = \mathop {\lim }\limits_{h \to 0} \left( {\dfrac{1}{h}\left( {\dfrac{1}{{\cos \left( {x + h} \right)}} - \dfrac{1}{{\cos x}}} \right)} \right) \\
   \Rightarrow \dfrac{d}{{dx}}\left( {\sec x} \right) = \mathop {\lim }\limits_{h \to 0} \left( {\dfrac{1}{h}\left( {\dfrac{{\cos x - \cos \left( {x + h} \right)}}{{\cos x.\cos \left( {x + h} \right)}}} \right)} \right) \\
\]
Now using trigonometric identity, $\cos A - \cos B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{B - A}}{2}} \right)$
\[
   \Rightarrow \dfrac{d}{{dx}}\left( {\sec x} \right) = \mathop {\lim }\limits_{h \to 0} \left( {\dfrac{1}{h}\left( {\dfrac{{2\sin \left( {\dfrac{{2x + h}}{2}} \right)\sin \left( {\dfrac{{x + h - x}}{2}} \right)}}{{\cos x.\cos \left( {x + h} \right)}}} \right)} \right) \\
   \Rightarrow \dfrac{d}{{dx}}\left( {\sec x} \right) = \mathop {\lim }\limits_{h \to 0} \left( {\dfrac{1}{h}\left( {\dfrac{{2\sin \left( {x + \dfrac{h}{2}} \right)\sin \left( {\dfrac{h}{2}} \right)}}{{\cos x.\cos \left( {x + h} \right)}}} \right)} \right) \\
    \\
 \]
Now, we use $\mathop {\lim }\limits_{x \to a} f\left( a \right) \times g\left( a \right) = \mathop {\lim }\limits_{x \to a} f\left( a \right) \times \mathop {\lim }\limits_{x \to a} g\left( a \right)$ where $\mathop {\lim }\limits_{x \to a} f\left( a \right)$ give finite value.
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\sec x} \right) = \mathop {\lim }\limits_{h \to 0} \left( {\dfrac{{\sin \left( {x + \dfrac{h}{2}} \right)}}{{\cos x.\cos \left( {x + h} \right)}}} \right) \times \mathop {\lim }\limits_{h \to 0} \left( {\dfrac{{\sin \left( {\dfrac{h}{2}} \right)}}{{\dfrac{h}{2}}}} \right)\]
We know, $\mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\sin x}}{x}} \right) = 1$
\[
   \Rightarrow \dfrac{d}{{dx}}\left( {\sec x} \right) = \mathop {\lim }\limits_{h \to 0} \left( {\dfrac{{\sin \left( {x + \dfrac{h}{2}} \right)}}{{\cos x.\cos \left( {x + h} \right)}}} \right) \times 1 \\
   \Rightarrow \dfrac{d}{{dx}}\left( {\sec x} \right) = \left( {\dfrac{{\sin \left( x \right)}}{{\cos x.\cos x}}} \right) \\
   \Rightarrow \dfrac{d}{{dx}}\left( {\sec x} \right) = \sec x.\tan x \\
 \]
So, the differentiation of $\sec x$ by using the first principle is \[\sec x.\tan x\].

Note-In such types of questions we generally face problems to solve limits so we use some important points. First we use the trigonometric identities to convert cosine into sine form because we know the most useful identity of limit is $\mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\sin x}}{x}} \right) = 1$ . So, after using this property we will get the required answer.