Last updated on July 4th, 2025
We use the derivative of tan(x), which is sec2(x), as a measuring tool for how the tangent function changes in response to a slight change in x. Derivatives help us calculate profit or loss in real-life situations. We will now talk about the derivative of tan(x) in detail.
We now understand the derivative of tan x. It is commonly represented as d/dx (tan x) or (tan x)', and its value is sec²x. The function tan x has a clearly defined derivative, indicating it is differentiable within its domain.
The key concepts are mentioned below:
Tangent Function: (tan(x) = sin(x)/cos(x)).
Quotient Rule: Rule for differentiating tan(x) (since it consists of sin(x)/cos(x)).
Secant Function: sec(x) = 1/cos(x).
The derivative of tan x can be denoted as d/dx (tan x) or (tan x)'. The formula we use to differentiate tan x is:
d/dx (tan x) = sec2 x (or)
(tan x)' = sec2 x
The formula applies to all x where cos(x) ≠ 0
We can derive the derivative of tan x using proofs. To show this, we will use the trigonometric identities along with the rules of differentiation. There are several methods we use to prove this, such as:
We will now demonstrate that the differentiation of tan x results in sec²x using the above-mentioned methods:
The derivative of tan x can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient.
To find the derivative of tan x using the first principle, we will consider f(x) = tan x. Its derivative can be expressed as the following limit.
f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1)
Given that f(x) = tan x, we write f(x + h) = tan (x + h).
Substituting these into equation (1),
f'(x) = limₕ→₀ [tan(x + h) - tan x] / h
= limₕ→₀ [ [sin (x + h) / cos (x + h)] - [sin x / cos x] ] / h
= limₕ→₀ [ [sin (x + h ) cos x - cos (x + h) sin x] / [cos x · cos(x + h)] ]/ h
We now use the formula sin A cos B - cos A sin B = sin (A - B).
f'(x) = limₕ→₀ [ sin (x + h - x) ] / [ h cos x · cos(x + h)]
= limₕ→₀ [ sin h ] / [ h cos x · cos(x + h)]
= limₕ→₀ (sin h)/ h · limₕ→₀ 1 / [cos x · cos(x + h)]
Using limit formulas, limₕ→₀ (sin h)/ h = 1.
f'(x) = 1 [ 1 / (cos x · cos(x + 0))] = 1/cos2 x
As the reciprocal of cosine is secant, we have,
f'(x) = sec2 x.
Hence, proved.
To prove the differentiation of tan x using the chain rule,
We use the formula:
Tan x = sin x/ cos x
Consider f(x) = sin x and g (x)= cos x
So we get, tan x = f (x)/ g(x)
By quotient rule: d/dx [f(x) / g(x)] = [f '(x) g(x) - f(x) g'(x)] / [g(x)]2… (1)
Let’s substitute f(x) = sin x and g (x) = cos x in equation (1),
d/ dx (tan x) = [(cos x) (cos x)- (sin x) (- sin x)]/ (cos x)2
(cos2 x + sin2 x)/ cos2 x …(2)
Here, we use the formula:
(cos2 x) + (sin2 x) = 1 (Pythagorean identity)
Substituting this into (2),
d/dx (tan x) = 1/ (cos x)2
Since sec x = 1/cos x, we write:
d/dx(tan x) = sec2 x
We will now prove the derivative of tan x using the product rule. The step-by-step process is demonstrated below:
Here, we use the formula,
Tan x = sin x/ cos x
tan x = (sin x). (cos x)-1
Given that, u = sin x and v = (cos x)-1
Using the product rule formula: d/dx [u.v] = u'. v + u. v'
u' = d/dx (sin x) = cos x. (substitute u = sin x)
Here we use the chain rule:
v = (cos x)-1 = (cos x)-1 (substitute v = (cos x)-1)
⇒ v' = -1. (cos)-2. d/dx (cos x)
v' = sin x/ (cos x)2
Again, use the product rule formula:
d/dx (tan x) = u'. v + u. V'
Let’s substitute u = sin x, u' = cos x, v = (cos x)-1, and v' = sin x/ (cos x)2
When we simplify each term:
We get, d/dx (tan x) = 1 + sin2x / (cos x)2
Sin2 x/ (cos x)2 = tan2 x (we use the identity sin2 x + cos2 x =1)
Thus:
d/dx (tan x) = 1 + tan2 x
Since, 1 + tan2 x = sec2 x
⇒ d/dx (tan x) = sec2 x.
When a function is differentiated several times, the derivatives obtained are referred to as higher-order derivatives. Higher-order derivatives can be a little tricky. To understand them better, think of a car where the speed changes (first derivative) and the rate at which the speed changes (second derivative) also changes. Higher-order derivatives make it easier to understand functions like tan (x).
Special Cases:
When the x is π/2, the derivative is undefined because tan (x) has a vertical asymptote there.
When the x is 0, the derivative of tan x = sec2 (0), which is 1.
Students frequently make mistakes when differentiating tan x. These mistakes can be resolved by understanding the proper solutions. Here are a few common mistakes and ways to solve them:
Calculate the derivative of (tan x·sec² x)
Here, we have f(x) = tan x·sec²x.
Using the product rule,
f'(x) = u′v + uv′
In the given equation, u = tan x and v = sec2 x.
Let’s differentiate each term,
u′= d/dx (tan x) = sec2 x
v′= d/dx (sec2 x) = 2 sec2 x tan x
substituting into the given equation,
⇒ f'(x) = (sec2 x). (sec2 x) + (tan x). (2 sec2 x tan x)
Let’s simplify terms to get the final answer,
f'(x) = sec4 x + 2 sec2 x tan2 x
Thus, the derivative of the specified function is sec4 x + 2 sec2 x tan2 x.
We find the derivative of the given function by dividing the function into two parts. The first step is finding its derivative and then combining them using the product rule to get the final result.
AXB International School sponsored the construction of a road. The slope is represented by the function y = tan(x) where y represents the elevation of the road at a distance x. If x = π/4 meters, measure the slope of the road.
We have y = tan (x) (slope of the road)...(1)
Now, we will differentiate the equation (1)
Take the derivative tan(x):
dy/ dx = sec2(x)
We know that, sec2 (x) = 1 + tan2 (x)
Given x = π/4 (substitute this into the derivative)
sec2 (π/4) = 1 + tan2 (π/4)
sec2 (π/4) = 1 + 12 = 2 (since tan (π/4) = 1)
Hence, we get the slope of the road at a distance x= π/4 as 2.
We find the slope of the road at x= π/4 as 2, which means that at a given point, the height of the road would rise at a rate twice the horizontal distance.
Derive the second derivative of the function y = tan (x).
The first step is to find the first derivative,
dy/dx = sec2 (x)...(1)
Now we will differentiate equation (1) to get the second derivative:
d2y/ dx2 = d/dx [sec2 (x)]
Here we use the product rule,
d2y / dx2 = 2 sec (x). d/dx [sec (x)]
d2y / dx2 = 2 sec (x). [ sec (x) tan (x)]
⇒ 2 sec2(x) tan (x)
Therefore, the second derivative of the function y = tan (x) is 2 sec2(x) tan (x).
We use the step-by-step process, where we start with the first derivative. Using the product rule, we differentiate sec2(x). We then substitute the identity and simplify the terms to find the final answer.
Prove: d/dx (tan² (x)) = 2 tan(x) sec² (x).
Let’s start using the chain rule:
Consider y = tan2 (x)
⇒ [tan (x)]2
To differentiate, we use the chain rule:
dy/dx = 2 tan (x). d/dx [tan (x)]
Since the derivative of tan (x) is sec2 (x),
dy/dx = 2 tan (x). sec2(x)
Substituting y = tan2(x),
d/dx (tan2(x)) = 2 tan (x). sec2(x)
Hence proved.
In this step-by-step process, we used the chain rule to differentiate the equation. Then, we replace tan (x) with its derivative. As a final step, we substitute y = tan2(x) to derive the equation.
Solve: d/dx (tan x/x)
To differentiate the function, we use the quotient rule:
d/dx (tan x/x) = (d/dx (tan x). X - tan x. d/dx(x))/ x2
We will substitute d/dx (tan x) = sec2x and d/dx (x) = 1
(sec2 x. x – tan x .1) / x2
= (x sec2 x – tan x) / x2
= x sec2 x – tan x / x2
Therefore, d/dx (tan x/x) = x sec2 x – tan x / x2
In this process, we differentiate the given function using the product rule and quotient rule. As a final step, we simplify the equation to obtain the final result.