Question

Find: $ \dfrac{d}{dx}(\sec x) $

Answer 1

Hint: Recall that sec x = $ \dfrac{1}{\cos x} $ = $ {{(\cos x)}^{-1}} $ and use the chain rule of derivatives: $ \dfrac{d}{dx}f[g(x)] $ = $ \dfrac{d}{d\ g(x)}f[g(x)] $ × $ \dfrac{d}{dx}g(x) $ . Knowing the derivatives of algebraic functions and trigonometric functions will be useful: $ \dfrac{d}{dx}{{x}^{n}} $ = $ n{{x}^{n-1}} $ and $ \dfrac{d}{dx}(\cos x) $ = −sin x.
We can also directly use the quotient rule of the derivatives $ \dfrac{d}{dx}(uv) $ = $ \dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}} $ for finding the derivative $ \dfrac{d}{dx}\left( \dfrac{1}{\cos x} \right) $ , taking the functions u = 1 and v = cos x.

Complete step-by-step answer:
We know that sec x = $ \dfrac{1}{\cos x} $ = $ {{(\cos x)}^{-1}} $ . Therefore, the given question $ \dfrac{d}{dx}(\sec x) $ can be written as $ \dfrac{d}{dx}{{(\cos x)}^{-1}} $ .
Using the chain rule of derivatives, we can write it as:
= $ \dfrac{d}{d(\cos x)}{{(\cos x)}^{-1}} $ × $ \dfrac{d}{dx}(\cos x) $
Applying the formulae $ \dfrac{d}{dx}{{x}^{n}} $ = $ n{{x}^{n-1}} $ and $ \dfrac{d}{dx}(\cos x) $ = −sin x, we will get:
= $ (-1){{(\cos x)}^{-1-1}} $ × (−sin x)
= $ \dfrac{\sin x}{{{\cos }^{2}}x} $ , which is the required answer.
It can also be written as:
= $ \dfrac{\sin x}{\cos x} $ × $ \dfrac{1}{\cos x} $
= tan x sec x, which is a more compact form of the same result.

Note: A quick observation will also tell us that $ \dfrac{d}{dx}(\csc x) $ = cot x csc x.
The derivatives of sin x and cos x are found using the first principle (definition) of derivatives. The first principle states that the derivative of a function f(x) is: $ \dfrac{d}{dx}f(x) $ = $ \underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(x+h)-f(x)}{h} $ . The derivative of a function is the rate of change of the value of the function with respect to the change in the value of the independent variable x. The derivative of a function y = f(x) is often also denoted by y' = f'(x).
It should be noted that integration is the opposite operation of differentiation.
That is, $ \int{f'(x)dx} $ = f(x) + C.