Sec x is the derivative of tan x. Let’s review 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). This is used to differentiate between tan and 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.

**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 sec^{2}x. Tan x is differentiable in its domain. To establish the differentiation of tan x to be sec^{2}x, we use the existing trigonometric identities and existing rules of differentiation. We can prove this in the following methods:

- Proof by the 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) = sec**^{2}x (or)**(tan x)’ = sec**^{2}x

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/cos^{2}x

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

f'(x) = sec^{2}x.

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) = -csc^{2}x. Also, by a property of exponents, a^{-m} = 1/a^{m}.

y’ = -1/cot^{2}x · (-csc^{2}x)

y’ = tan^{2}x · csc^{2}x

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

y’ = (sin^{2}x)/(cos^{2}x) · (1/sin^{2}x)

= 1/cos^{2}x

We have 1/cos x = sec x. So

y’ = sec^{2}x

Hence proved.

**Derivative of Tan x Proof by Quotient Rule**

We can evaluate 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)] / (cos^{2}x)

= [cos x · cos x – sin x (-sin x)] / (cos^{2}x)

= [cos^{2}x + sin^{2}x] / (cos^{2}x)

By one of the Pythagorean identities, cos^{2}x + sin^{2}x = 1. So

y’ = 1 / (cos^{2}x) = sec^{2}x

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

**Common Misconceptions Related to Derivative of Tan x:**

Here is some clarity concerning 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
^{-1}x are NOT same.

d/dx(tan x) = sec^{2}x

d/dx(tan^{-1}x) = 1/(1 + x^{2})

**FAQs**

**Q: What is the Derivative of Tan x with Respect to x?**

A; The derivative of tan x with respect to x is the square of sec x. i.e., d/dx(tan x) = sec^{2}x. It can also be written as (tan x)’ = sec^{2}x.

**Q: What is the Derivative of Tan x ^{2}?**

A: We know that d/dx(tan x) = sec^{2}x. So d/dx(tan x^{2}) = sec^{2}x^{2} d/dx(x^{2}) = 2xsec^{2}x^{2 }(by chain rule).

**Q: What is the Differentiation of Tan x in Terms of Cos x?**

A: We know that the derivative of tan x is sec^{2}x. Also, sec x = 1/(cos x). So d/d(tan x) = 1/cos^{2}x.

**Q: What is the Derivative of tan x ^{-1}?**

A: By using the derivative of tan x and chain rule, d/dx(tan x^{-1}) = sec^{2}x^{-1} d/dx(x^{-1}) = sec^{2}x^{-1} (-1 x^{-2}) = (-sec^{2}x^{-1})/(x^{2}).

**Q: Is the Derivative of Tan x Equal to Derivativative of Tan ^{-1}x?**

A; No, the derivatives of tan x and tan^{-1}x are different. The derivative of tan x is sec^{2}x whereas the derivative of tan⁻¹x is 1/(1 + x^{2}).

**Q: What is the Difference Between the Derivative of Tan x and the Antiderivative of Tan x?**

A: The derivative of tan x is sec^{2}x. The antiderivative of tan x is nothing but the integral of tan x and ∫ tan x dx = ln |sec x| + C.

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