Derivative of `sec(x)`
- Introduction
- What Do You Mean by Derivative of `sec x`?
- Proving Derivative of `sec x` Using First Principle
- Proving Derivative of `sec x` Using Chain Rule
- Proving Derivative of `sec x` Using Quotient Rule
- Solved Examples
- Practice Problems
- Frequently Asked Questions
Introduction
In math, the derivative of a function `f(x)` is the rate of change of the function with respect to its independent variable `x`. The derivative of a function `f(x)` is represented by `f'(x)` or ` \frac{d}{dx}f(x)`. The derivative of a trigonometric function is also called the differentiation of the trigonometric function.
Prior to determining the derivative of `sec x`, let's understand that `sec x` is the reciprocal of the basic trigonometric function `cos x`, while `tan x` is the ratio of `sin x` to `cos x`. To differentiate `sec x` with regard to `x`, it is crucial to understand the meanings of `sec x` and `tan x`.
What Do You Mean by Derivative of `sec x`?
Among many of the trigonometric derivatives, the derivative of `sec x` is one of the derivatives. The derivative of `sec x` with respect to `x` is `sec x · tan x`. The derivative of `sec x` is the rate of change of the function `sec x` with respect to the angle `x`. We use `d/dx(sec x)` (or) `(sec x)'` to indicate the derivative of `sec x` with respect to `x`. Consequently,
If `(sec x)' = (sec x)· tan x`, then `d/dx (sec x) = sec x · tan x`.
Proving Derivative of `sec x` Using First Principle
The derivative of `sec x` can be proved using the following ways:
- By using the First Principle of Derivative
- By using Quotient Rule
- By using Chain Rule
To find the derivative of `\sec(x)` using the first principles (also known as the definition of the derivative), we'll apply the limit definition of the derivative:
`f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}`
For `f(x) = \sec(x)`, let's proceed with the calculation:
`f(x) = \sec(x) = \frac{1}{\cos(x)}`
So, we need to evaluate:
`f'(x) = \lim_{h \to 0} \frac{\frac{1}{\cos(x + h)} - \frac{1}{\cos(x)}}{h}`
`= \lim_{h \to 0} \frac{\cos(x) - \cos(x + h)}{h \cdot \cos(x) \cdot \cos(x + h)}`
Now, we can apply the trigonometric identity ` \cos(a) - \cos(b) = 2 \sin\left(\frac{a+b}{2}\right)\sin\left(\frac{b-a}{2}\right) `:
`= \lim_{h \to 0} \frac{2 \sin\left(\frac{x + (x + h)}{2}\right)\sin\left(\frac{(x + h)- x }{2}\right)}{h \cdot \cos(x) \cdot \cos(x + h)}`
`= \lim_{h \to 0} \frac{2 \sin\left(x + \frac{h}{2}\right)\sin\left(\frac{h}{2}\right)}{h \cdot \cos(x) \cdot \cos(x + h)}`
Now, as ` h ` approaches zero, ` \frac{h}{2} ` also approaches zero. So, we can write:
`= \lim_{h \to 0} \frac{\sin\left(x + \frac{h}{2}\right)}{\cdot \cos(x) \cdot \cos(x + h)}\lim_{\frac{h}{2} \to 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}}`
Since ` lim_{x \to 0} \frac{sin x}{x} = 1 `, we can write
`f'(x) = \frac{sin x}{cos x \cdot cos x} \times 1`
`f'(x) = sec x \cdot tan x`
So, the derivative of ` \sec(x) ` using the first principle is ` sec \cdot tan x `.
Proving Derivative of `sec x` Using Chain Rule
To find the derivative of ` \sec(x) ` using the chain rule, we start by expressing ` \sec(x) ` as ` \sec(x) = \frac{1}{\cos x} = (\cos x)^{-1}`.
Now, applying the power rule and chain rule:
`\frac{d}{dx}(\sec(x)) = -\frac{1}{\cos^2(x)} \cdot (-\sin(x))`
`= \frac{\sin(x)}(\cos^2(x))`
`= \frac{\sin(x)}{\cos(x) \cdot \cos(x)}`
`= \frac{\sin(x)}\cos(x) \cdot \frac{1}\cos(x)`
`= \tan(x) \cdot \sec(x)`
So, the derivative of ` \sec(x) ` with respect to ` x ` using the chain rule is ` \tan(x) \cdot \sec(x) `.
Proving Derivative of `sec x` Using Quotient Rule
To find the derivative of ` \sec(x) ` using the quotient rule, we start by expressing ` \sec(x) ` as ` \sec(x) = \frac{1}{\cos(x)} `. Then, we apply the quotient rule, which states that if we have a function of the form ` \frac{u(x)}{v(x)} `, then its derivative is given by:
`\frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{[v(x)]^2}`
Let `u(x) = 1` and `v(x) = \cos(x)`.
Then, we have ` \sec(x) = \frac{u(x)}{v(x)} `.
Now, we differentiate ` u(x) = 1 ` and ` v(x) = \cos(x) ` with respect to ` x `:
\( u'(x) = 0 \)
\( v'(x) = -\sin(x) \)
Now, applying the quotient rule:
`\frac{d}{dx}\left(\frac{1}{\cos(x)}\right) = \frac{\cos(x) \cdot 0 - 1 \cdot (-\sin(x))}{\cos^2(x)}`
`= \frac{\sin(x)}{\cos^2(x)}`
`= \frac{\sin(x)}{\cos(x) \cdot \cos(x)}`
`= \frac{\sin(x)}{\cos(x)} \cdot \frac{1}{\cos(x)}`
`= \tan(x) \cdot \sec(x)`
So, the derivative of ` \sec(x) ` with respect to ` x ` using the quotient rule is ` \tan(x) \cdot \sec(x) `.
Solved Examples
Example `1`: Find the secant derivative of `7x` with respect to `x`, that is ` \frac{dy}{dx} \sec(7x)`.
Solution:
To find the derivative of ` \sec(7x) ` with respect to ` x `, we'll use the chain rule. The derivative of ` \sec(u) ` with respect to ` u ` is ` \sec(u) \tan(u) `, and then we multiply by the derivative of the inner function `(7x)` with respect to ` x `, which is `7`.
So, applying the chain rule, we have:
`\frac{d}{dx}[\sec(7x)] = \sec(7x) \tan(7x) \cdot 7`
`\frac{d}{dx}[\sec(7x)] = 7\sec(7x) \tan(7x)`
That's the derivative of ` \sec(7x) ` with respect to ` x `.
Example `2`: Find the derivative of ` y = 2\sec(x) ` with respect to ` x `.
Solution:
Given function: ` y = 2\sec(x) `
Using the chain rule, the derivative is:
`\frac{dy}{dx} = 2 \cdot \tan(x) \cdot \sec(x)`
So, the derivative of ` 2\sec(x) ` with respect to ` x ` is ` 2 \tan(x) \sec(x) `.
Example `3`: Find the derivative of ` y = \sec^2(x) ` with respect to ` x `.
Solution:
Given function: ` y = \sec^2(x) `
Using the chain rule, the derivative is:
`\frac{dy}{dx} = 2\sec(x) \cdot \sec(x) \tan(x) = 2\sec^2(x)\tan(x)`
So, the derivative of ` 2\sec^2(x) ` with respect to ` x ` is ` 2\sec^2(x)\tan(x) `.
Example `4`: Find the derivative of ` y = \sec(x)\tan(x) ` with respect to ` x `.
Solution:
The function we are considering is
\( f(x) = \sec(x) \tan(x) \)
To find the derivative of this function, `f'(x)`, we can use the product rule. Recall that the product rule states that if you have two functions `u(x)` and `v(x)`, then the derivative of their product is given by
\( (uv)' = u'v + uv' \)
Here, let
\( u(x) = \sec(x), \quad v(x) = \tan(x) \)
We need the derivatives of each:
\( u'(x) = \sec(x)\tan(x) \quad \text{(derivative of secant)} \)
\( v'(x) = \sec^2(x) \quad \text{(derivative of tangent)} \)
Applying the product rule, we get:
\( f'(x) = u'(x)v(x) + u(x)v'(x) = \sec(x)\tan(x)\tan(x) + \sec(x)\sec^2(x) \)
\( f'(x) = \sec(x)\tan^2(x) + \sec^3(x) \)
Thus, the derivative of the function `\sec(x)\tan(x)` is
\( {\sec(x)(\tan^2(x) + \sec^2(x))} \)
Practice Problems
Q`1`. What is the derivative of ` \sec(3x) ` with respect to ` x `?
- ` 3\sec(3x) \tan(3x) `
- ` \sec(3x) \tan(3x) `
- ` 3\csc(x) \cot(x) `
- ` -3\sec(3x) \tan(3x) `
Solution: a
Q`2`. If ` y = \sec(x) `, what is ` \frac{dy}{dx} `?
- ` \sec(x) \tan(x) `
- ` \csc(x) \cot(x) `
- ` -\sec(x) \tan(x) `
- ` -\csc(x) \cot(x) `
Solution: a
Q`3`. What is the derivative of ` \sec^2(x) ` with respect to ` x `?
- ` -2\sec(x) \tan(x) `
- ` 2\sec(x) \tan^2(x) `
- ` 2\sec(x) `
- ` 2\sec^2(x) \tan(x) `
Solution: d
Q`4`. What is the derivative of ` \frac{1}{\sec(x)} ` with respect to ` x `?
- ` \frac{\sin(x)}{\cos^2(x)} `
- ` -\sec(x) \tan(x) `
- ` -\sin(x) `
- ` \sec(x) \tan(x) `
Solution: c
Q`5`. If ` y = 5 \sec(2x) `, what is ` \frac{dy}{dx} `?
- ` 10\sec(2x) \tan(2x) `
- ` 5\sec(2x) \tan(2x) `
- ` \frac{1}{2} \sec(2x) \tan(2x) `
- ` \frac{1}{2} \sec(2x) `
Solution: a
Frequently Asked Questions
Q`1`. What is the derivative of ` \sec(x) ` with respect to ` x `?
Answer: The derivative of ` \sec(x) ` with respect to ` x ` is ` \tan(x) \cdot \sec(x) `.
Q`2`. How do you find the derivative of ` \sec(x) ` using the chain rule?
Answer: To find the derivative of ` \sec(x) ` using the chain rule, you first express ` \sec(x) ` as ` \frac{1}{\cos(x)} `, then differentiate using the chain rule with ` f(u) = \frac{1}{u} ` and ` g(x) = \cos(x) `.
Q`3`. What is the derivative of ` y = \sec^2(x) ` with respect to ` x `?
Answer: The derivative of ` y = \sec^2(x) ` with respect to ` x ` is ` 2\sec^2(x)\tan(x) `.
Q`4`. Can you explain the steps involved in finding the derivative of ` \frac{\sec(x)}{x} `?
Answer: To find the derivative of ` \frac{\sec(x)}{x} `, you use the quotient rule along with the chain rule.
Q`5`. How do you apply the quotient rule to find the derivative of a function involving ` \sec(x) `?
Answer: When applying the quotient rule to find the derivative of a function involving ` \sec(x) ` we consider ` \sec(x) ` as ` \frac{1}{\cos(x)} ` , differentiate the numerator and denominator separately, and then apply the quotient rule formula: `\frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{[v(x)]^2}`.