Derivative Of Tan X - Formula, Proof, Examples - Cuemath

Derivative of Tan x

The derivative of tan x is the square of sec x. Before proving this, let us recollect some facts about tan x. Tan x in a right-angled triangle is the ratio of the opposite side of x to the adjacent side of x and thus it can be written as (sin x)/(cos x). We use this in doing the differentiation of tan x.

Let us learn the differentiation of tan x along with its proof in different methods and also we will solve a few examples using the derivative of tan x.

1.	What is the Derivative of Tan x?
2.	Derivative of Tan x Proof by First Principle
3.	Derivative of Tan x Proof by Chain Rule
4.	Derivative of Tan x Proof by Quotient Rule
5.	FAQs on Derivative of Tan x

What is the Derivative of Tan x?

The derivative of tan x with respect to x is denoted by d/dx (tan x) (or) (tan x)' and its value is equal to sec2x. Tan x is differentiable in its domain. To prove the differentiation of tan x to be sec2x, we use the existing trigonometric identities and existing rules of differentiation. We can prove this in the following ways:

• Proof by first principle
• Proof by chain rule
• Proof by quotient rule

Derivative of Tan x Formula

The formula for differentiation of tan x is,

• d/dx (tan x) = sec2x (or)
• (tan x)' = sec2x

Now we will prove this in different methods in the upcoming sections.

Derivative of Tan x Proof by First Principle

To find the derivative of tan x, we assume that f(x) = tan x. Then by first principle, its derivative is given by the following limit.

f'(x) = limₕ→₀ [f(x + h) - f(x)] / h ... (1)

Since f(x) = tan x, we have f(x + h) = tan (x + h).

Substituting these in (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

By sum and difference formulas, 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)]

By limit formulas, limₕ→₀ (sin h)/ h = 1.

f'(x) = 1 [ 1 / (cos x · cos(x + 0))] = 1/cos2x

We know that the reciprocal of cos is sec. So,

f'(x) = sec2x.

Hence proved.

Derivative of Tan x Proof by Chain Rule

We will prove the differentiation of tan x formula by chain rule. For this let us note that we can write y = tan x as y = 1 / (cot x) = (cot x)-1. Now, by power rule and chain rule,

y' = (-1) (cot x)-2 · d/dx (cot x)

We have d/dx (cot x) = -csc2x. Also, by a property of exponents, a-m = 1/am.

y' = -1/cot2x · (-csc2x)

y' = tan2x · csc2x

Now, tan x = (sin x)/(cos x) and csc x = 1/(sin x). So

y' = (sin2x)/(cos2x) · (1/sin2x)

= 1/cos2x

We have 1/cos x = sec x. So

y' = sec2x

Hence proved.

Derivative of Tan x Proof by Quotient Rule

We can apply the quotient rule to derive the formula of the derivative of tan x. For this, we have to write 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)] / (cos2x)

= [cos x · cos x - sin x (-sin x)] / (cos2x)

= [cos2x + sin2x] / (cos2x)

By one of the Pythagorean identities, cos2x + sin2x = 1. So

y' = 1 / (cos2x) = sec2x

Hence proved. This proof is the easiest one among all the other proofs of the derivatives of tan x.

Common Misconceptions Related to Derivative of Tan x:

Here is some clarity about some common misconceptions regarding the differentiation of tan x.

• d/dx (tan x) is NOT equal to d/dx(sin x) / d/dx (cos x). Instead, we have to use the quotient rule to find the derivative of tan x (by writing it as (sin x)/(cos x)).
• d/dx (tan x) is NOT cot x. cot x is just the reciprocal of tan x.
• The derivatives of tan x and tan-1x are NOT same. d/dx(tan x) = sec2x d/dx(tan-1x) = 1/(1 + x2)

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