In this article, you will learn what is the derivative of tan x as well as prove the derivative of tan x by chain rule, first principal rule, and quotient rule.
I have already discussed the derivative of sin x, cos x, and sec x in a previous article you can check from here.
So, without wasting time let's get started.
What is tan x?
tan x is a trigonometric function which is equal to the (sin x)/(cos x).
This trigonometric function is mostly used in right-angle triangles.
Derivative of tan x
The derivative of tan x is equal to the sec²x.
We can prove the derivative of tan x in three ways first by using the quotient rule and second by using the first principle rule and the last chain rule.
Derivative of tan x Proof by Quotient Rule
The formula of the quotient rule is,
dy/dx = {v (du/dx) - u (dv/dx)}/v²
Where,
dy/dx = derivative of y with respect to x
v = variable v
du/dx = derivative of u with respect to x
u = variable u
dv/dx = derivative of v with respect to x
v = variable v
As we know,
tan x = sin x/cos x
So,
Let us
y = tan x
and,
u = sin x
v = cos x
Now putting these values on the quotient rule formula, we will get
dy/dx = [ cos x × d/dx (sin x) - sin x × d/dx (cos x)] / (cos²x)
= [cos x · cos x - sin x (-sin x)] / (cos²x)
= [cos²x + sin²x] / (cos²x)
So, from the Pythagoras theorem, we know that cos²x + sin²x = 1
So,
dy/dx = 1 / cos²x
So, 1 / cos²x = sec²x
Hence,
d/dx(tan x) = sec²x
Thus, we proved the derivative of tan x will be equal to sec²x using the quotient rule method.
Derivative of tan x Proof by First Principle Rule
According to the first principle rule, the derivative limit of a function can be determined by computing the formula:
For a differentiable function y = f (x)
We define its derivative w.r.t x as :
dy/dx = f ' (x) = limₕ→₀ [f(x+h) - f(x)]/h
f'(x) = limₕ→₀ [f(x+h) - f(x)]/h
This limit is used to represent the instantaneous rate of change of the function f(x).
Let,
f (x) = tan x
So,
f(x + h) = tan (x + h)
Putting these values on the above first principle rules equation.
f'(x) = limₕ→₀ [tan(x + h) - tan x] / h
Since, we know tan x = sin x/cos x
So, putting these values
f'(x) = 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
As we know,
sin (a - b) = sin a cos b - cos a sin b
So,
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)]
So, after applying the limit,
limₕ→₀ (sin h)/ h = 1.
f'(x) = 1 [ 1 / (cos x × cos(x + 0))]
f'(x) = 1/cos² x
As we know,
1/cos x = sec x
So,
f'(x) = sec²x.
Thus, we proved the derivative of tan x will be equal to sec²x using the first principle rule method.
Derivative of tan x Proof by Chain Rule
Let us,
y = tan x
As we know,
tan x = 1/cot x
So,
y = 1 / (cot x)
= (cot x)⁻¹
By using the chain rule,
The formula of chain rule is,
dy/dx = (dy/du) × (du/dx)
Where,
dy/dx = derivative of y with respect to x
dy/du = derivative of y with respect to u
du/dx = derivative of u with respect to x
After putting these values we can find,
dy/dx = -1 (cot x)⁻² × d/dx (cot x)
Since,
d/dx (cot x) = (-cosec²x) and a⁻ⁿ = 1/aⁿ
So,
dy/dx = -1/(cot² x) × (-cosec²x)
As we know,
1/(cot² x) = tan² x
So,
dy/dx = (tan² x) × (cosec²x)
Now we know that the trigonometric formula
tan x = sin x/cos x and cosec x = 1/sin x
So after putting these values we will get,
dy/dx = (sin² x)/(cos² x) × (1/sin² x)
= 1/cos² x
As we know,
1/cos x = sec x
d/dx (tan x) = sec²x
Thus, we proved the derivative of tan x will be equal to sec²x using the chain rule method.
See Also:
FAQ Related to Derivative of tan x
What is a derivative of tan x?
The derivative of tan x is equal to the sec²x.
Where is tan equal to 1?
At value tan 45° is equal to 1.
What is tan Infinity?
At value tan 90° is equal to infinity.
What is the differentiation of tan x?
The differentiation of tan x is equal to the sec²x.
So friends here I discussed all aspects related to the derivative of tan x.
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