Question

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

Answer 1

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.