Question

Prove that the
 $ \dfrac{d}{{dx}}(\sec x) = \sec x\tan x $

Answer 1

Hint: The question asks us to prove a standard formula we use in the differentiation of the secant functions. The formula is
 $ \dfrac{d}{{dx}}(\sec x) = \sec x\tan x $
The formula will be proved by using the quotient rule of differentiation. The formula is used for differentiation of the functions which have a denominator with a variable. The formula for differentiation of a function of the form $ \left( {\dfrac{f}{g}} \right) $ where $ f $ is the numerator and $ g $ is denominator is given below.
\[{\left( {\dfrac{f}{g}} \right)^\prime } = \dfrac{{f'{\mkern 1mu} g - f{\mkern 1mu} g'}}{{{g^2}}}\]
The notation of ‘ here refers to the differentiation.

Complete step-by-step answer:
The given formula will be proved on the basis of the product rule we will express the $ \sec x $ as
 $ \Rightarrow \sec x = \dfrac{1}{{\cos x}} $
And then do the differentiation using the quotient rule.
The numerator here is $ 1 $ whose differentiation will be $ 0 $ and the denominator here is $ \cos x $ whose differentiation will be $ - \sin x $ . So we will write quotient rule as,
 $ \Rightarrow {\left( {\dfrac{1}{{\cos x}}} \right)^\prime } = \dfrac{{(1)'{\mkern 1mu} \cos x - 1(\cos x)'}}{{{{(\cos x)}^2}}} $
 $ \Rightarrow {\left( {\dfrac{1}{{\cos x}}} \right)^\prime } = \dfrac{{0 - 1( - \sin x)}}{{{{(\cos x)}^2}}} $
 $ \Rightarrow {\left( {\dfrac{1}{{\cos x}}} \right)^\prime } = \dfrac{{\sin x}}{{{{(\cos x)}^2}}} $
The right hand side of the equation can be written as multiplication of two factors as below,
 $ \Rightarrow {\left( {\dfrac{1}{{\cos x}}} \right)^\prime } = \dfrac{1}{{\cos x}} \times \dfrac{{\sin x}}{{\cos x}} $
We know that $ \dfrac{1}{{\cos x}} = \sec x $
And also that $ \dfrac{{\sin x}}{{\cos x}} = \tan x $
We will thus write our equation as,
 $ \Rightarrow {\left( {\dfrac{1}{{\cos x}}} \right)^\prime } = \sec x\tan x $
Hence we have proved the formula for differentiation of the secant function. This formula is very important in differential calculus and hence should be remembered by heart.
So, the correct answer is “Option B”.

Note: The two important formulas that can help us in solving the differential calculus apart from the standard formulas of trigonometric ratios or formula for differentiation of $ {x^n} $ are the product and the quotient formula, The quotient formula has already been mentioned which can help us to solve any functions with a denominator. The other formula, the product formula, is useful in finding the differentiation of a function which is the product of two functions. It is written as below,
 $ (fg)' = f(g)' + (f)'g $
Remember here that the (‘) above the function means the differentiation of that function.