Derivative of `tan(x)`
- Introduction
- What do you understand by the derivative of `tan x`
- Proving Derivative of `tan x` Using First Principle
- Proving Derivative of `tan x` Using Chain Rule
- Proving Derivative of `tan x` Using Quotient Rule
- Solved Examples
- Practice Problems
- Frequently Asked Questions
Introduction
`Tan x`, which can be expressed as `(sin x)/(cos x)`, is the ratio of the opposite side of `x` to the adjacent side of `x` in a right-angled triangle. We use this to differentiate `tan x`. The derivative of `tan x` is the rate at which the function `tan x` changes with respect to the angle `x`.
The square of `sec x` is the derivative of `tan x`. Let's review some `tan x` facts before demonstrating this.
Let's study the differentiation of `tan x`, its various techniques of demonstration, and how to use the derivative of `tan x` to solve a few instances.
What Do You Understand by The Derivative of `tan x`
The derivative of the tangent function `tan(x)` is `sec^2(x)`, where `sec(x)` is the secant function. This indicates that the square of the secant function evaluated at `x` equals the rate of change of the tangent function with respect to `x`. It may be written as follows in mathematical notation:
\( \frac{d}{dx} \tan(x) = \sec^2(x) \)
Proving Derivative of `tan x` Using First Principle
In order to find the derivative of `tan x` using first principle, let us assume that `f(y) = tan y`. By using first principle, its derivative is given by
`f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \quad .... (1)`
Since `f(x) = tan x`, we have `f(x + h) = tan (x + h)`.
Substituting the values in equation `(1)`,
`f'(y) = \lim_{h \to 0} [tan(x+ h) - tan x] / h`
`= \lim_{h \to 0} [ [sin (x + h) / cos (x+ h)] - [sin x / cos x] ] / h`
`= \lim_{h \to 0} [ [sin (x + h ) cos x - cos (x + h) sin x] / [cos x · cos(x + h)] ]/ h`
By using the sum and difference formula, we get, `sin A cos B - cos A sin B = sin (A - B)`.
`f'(y) = \lim_{h \to 0} [ sin (x + h - x) ] / [ h cos x · cos(x + h)]`
`= \lim_{h \to 0} [ sin h ] / [ h cos x · cos(x + h)]`
`= \lim_{h \to 0} (sin h)/ h · \lim_{h \to 0} 1 / [cos x · cos(x + h)]`
By limit formulas, `\lim_{h \to 0} (sin h)/ h = 1`.
`f'(y) = 1 [ 1 / (cos x · cos(x + 0))] = 1/(cos^2x)`
We know that the reciprocal of `cos` is `sec`.
So, we have `f'(x) = sec^2x`.
So, the derivative of \( \tan(x) \) using the first principle is \( sec^2(x) \).
Proving Derivative of `tan x` Using Chain Rule
Let us assume `y = tan x` as `y = 1 / (cot x) = (cot x)^-1`.
By using power rule and chain rule,
`y' = (-1) (cot x)^-2 · d/dx (cot x)`
We have `d/dx (cot x) = -csc^2x`. Now by using the property of exponents,we have `a^-m = 1/(a^m)`.
`y' = -1/(cot^2x) · (-csc^2x)`
`y' = tan^2x · csc^2x`
Now, `tan x = (sin x)/(cos x)` and `csc x = 1/(sin x)`. So
`y' = (sin^2x)/(cos^2x) · (1/(sin^2x))`
`= 1/(cos^2x)`
We have `1/cos x = sec x`. So
`y' = sec^2x`
So, the derivative of \( \tan(x) \) using the chain rule is \( sec^2(x) \).
Proving Derivative of `tan x` Using Quotient Rule
We need to express `tan x` as a fraction. We know that `tan x = (sin x)/(cos x)`. So we assume that `y = (sin x)/(cos x)`. Then by quotient rule,
`y' = [ cos x · d/dx (sin x) - sin x · d/dx (cos x)] / (cos^2x)`
`= [cos x· cos x - sin x (-sin x)] / (cos^2x)`
`= [cos^2x + sin^2x] / (cos^2x)`
By one of the Pythagorean identities, `cos^2x + sin^2x = 1`. So
`y' = 1 / (cos^2x) = sec^2x`
So, the derivative of \( \tan(x) \) using the quotient rule is \( sec^2(x) \).
Solved Examples
Example `1`: Find the derivative of \( y = \tan(3x) \) using the chain rule.
Solution:
Given \( y = \tan(3x) \), we apply the chain rule, which states that if \( y = f(g(x)) \), then \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).
Let \( u = 3x \). Then, \( \frac{du}{dx} = 3 \).
Now, \( y = \tan(u) \), so \( \frac{dy}{du} = \sec^2(u) \).
Using the chain rule, we have:
\( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \sec^2(u) \cdot 3 = 3\sec^2(3x) \)
Therefore, the derivative of \( y = \tan(3x) \) is \( 3\sec^2(3x) \).
Example `2`: Find the derivative of \( y = \tan^2(x) \) using the chain rule.
Solution:
Given \( y = \tan^2(x) \), we rewrite \( \tan^2(x) \) as \( (\tan(x))^2 \).
Let \( u = \tan(x) \), so \( y = u^2 \).
Now, \( \frac{dy}{du} = 2u \).
Using the chain rule, we have:
\( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 2u \cdot \sec^2(x) = 2\tan(x) \cdot \sec^2(x) \)
Therefore, the derivative of \( y = \tan^2(x) \) is \( 2\tan(x) \cdot \sec^2(x) \).
Example `3`: Find the derivative of \( y = \frac{\tan(x)}{x} \) using the quotient rule.
Solution:
Given \( y = \frac{\tan(x)}{x} \), we apply the quotient rule:
\( \frac{d}{dx} \left(\frac{\tan(x)}{x}\right) = \frac{x \cdot \frac{d}{dx} \tan(x) - \tan(x) \cdot \frac{d}{dx} x}{x^2} \)
Recall that \( \frac{d}{dx} \tan(x) = \sec^2(x) \) and \( \frac{d}{dx} x = 1 \).
Substituting these derivatives into the quotient rule formula, we get:
\( \frac{x \cdot \sec^2(x) - \tan(x) \cdot 1}{x^2} = \frac{x \sec^2(x) - \tan(x)}{x^2} \)
Therefore, the derivative of \( y = \frac{\tan(x)}{x} \) is \( \frac{x \sec^2(x) - \tan(x)}{x^2} \).
Example `4`: Find the derivative of \( y = \sin(x) \tan(x) \) using the product rule.
Solution:
Given \( y = \sin(x) \tan(x) \), we apply the product rule:
\( \frac{d}{dx} (\sin(x) \tan(x)) = \sin(x) \frac{d}{dx} \tan(x) + \tan(x) \frac{d}{dx} \sin(x) \)
We know that \( \frac{d}{dx} \tan(x) = \sec^2(x) \) and \( \frac{d}{dx} \sin(x) = \cos(x) \).
Substituting these derivatives into the product rule formula, we get:
\( \sin(x) \sec^2(x) + \tan(x) \cos(x) \)
\( = \tan(x) \sec(x) + \tan(x) \cos(x) \)
\( = \tan(x) \sec(x) + \sin(x) \)
Therefore, the derivative of \( y = \sin(x) \tan(x) \) is \( \tan(x) \sec(x) + \sin(x) \).
Practice Problems
Q`1`. What is the derivative of \( \tan(x) \)?
- \( \sec(x) \)
- \( \csc(x) \)
- \( \cot(x) \)
- \( \sec^2(x) \)
Answer: d
Q`2`. If \( f(x) = \tan(3x) \), what is \( f'(x) \)?
- \( \cos^3(3x) \)
- \( 3\sec^2(3x) \)
- \( 3\tan^2(3x) \)
- \( 3\cos(3x) \)
Answer: b
Q`3`. What is the derivative of \( g(x) = \tan(x)\sec^2(x) \)?
- \( 2\tan(x) \)
- \( 2\sec^2(x)\tan^2(x) + \sec^4(x) \)
- \( 2\cot(x) \)
- \( 2\sec^2(x)\tan^2(x) \)
Answer: b
Q`4`. If \( h(x) = \tan(5x) \), what is \( h'(x) \)?
- \( 5\sec^2(5x) \)
- \( \sec^2(5x) \)
- \( \frac{5}{\sec^2(5x)} \)
- \( \frac{\sec^2(5x)}{5} \)
Answer: a
Q`5`. What is the derivative of \( f(x) = \cos(x) \tan(x) \)?
- \( \sin(x) \sec^2(x) \)
- \( \tan(x) \cos(x) \)
- \( \cos(x)\)
- \( \sin(x) \cos(x) \)
Answer: c
Frequently Asked Questions
Q`1`. What is the derivative of \( \tan(x) \)?
Answer: The derivative of \( \tan(x) \) with respect to \( x \) is \( \sec^2(x) \). This means that the rate of change of the tangent function at any point \( x \) is equal to the square of the secant function evaluated at \( x \).
Q`2`. How is the derivative of \( \tan(x) \) derived?
Answer: The derivative of \( \tan(x) \) can be derived using the quotient rule or by rewriting \( \tan(x) \) in terms of sine and cosine and then applying the chain rule. The final result is \( \sec^2(x) \).
Q`3`. Why is the derivative of \( \tan(x) \) equal to \( \sec^2(x) \)?
Answer: The derivative of \( \tan(x) \) is \( \sec^2(x) \) because \( \tan(x) \) is defined as \( \frac{\sin(x)}{\cos(x)} \). When finding the derivative, we apply differentiation rules and trigonometric identities to arrive at \( \sec^2(x) \).
Q`4`. What are the key properties of the derivative of \( \tan(x) \)?
Answer: Some key properties of the derivative of \( \tan(x) \) include:
- The derivative of \( \tan(x) \) is never undefined for real values of \( x \).
- The graph of \( \tan(x) \) has vertical asymptotes where \( \cos(x) = 0 \), affecting the behavior of its derivative near these points.
- The derivative of \( \tan(x) \) is positive in some intervals and negative in others, depending on the values of \( x \).
Q`5`. How do I find the derivative of composite functions involving \( \tan(x) \)?
Answer: When dealing with composite functions involving \( \tan(x) \), you can use the chain rule. For example, if you have \( f(x) = \tan(3x) \), you would differentiate \( \tan(3x) \) as \( \sec^2(3x) \) using the chain rule, then multiply by the derivative of the inner function, which is \( 3 \).