# Derivative of Tangent, tan(x) – Formula, Proof, and Graphs

The Derivative of Tangent is one of the first transcendental functions introduced in Differential Calculus (or Calculus I). The derivative of the tangent function is equal to secant squared, sec2(x). We can prove this derivative using limits and trigonometric identities.

In this article, we will discuss how to derive the trigonometric function tangent. We will cover brief fundamentals, its formula, a graph comparison of tangent and its derivative, a proof, methods to derive, and a few examples.

##### CALCULUS

Relevant for

Learning about the proof and the graphs of the derivative of tangent.

See proof

##### CALCULUS

Relevant for

Learning about the proof and the graphs of the derivative of tangent.

See proof

## Proof of the Derivative of the Tangent Function using limits

The trigonometric function tangent of an angle is defined as the ratio of the opposite side to the adjacent side of an angle in a right triangle. Illustrating it through a figure, we have

where C is 90°. Therefore, getting the tangent of angle A can be evaluated as

$latex \tan{(A)} = \frac{a}{b}$

Before learning the proof of the derivative of the tangent function, you are hereby recommended to learn the Pythagorean theorem, Soh-Cah-Toa & Cho-Sha-Cao, and the first principle of limits as prerequisites.

To review, any function can be derived by equating it to the limit of

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\frac{f(x+h)-f(x)}{h}}$$

Suppose we are asked to get the derivative of

$latex f(x) = \tan{(x)}$

we have

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\frac{ \tan{(x+h)} – \tan{(x)} }{h}}$$

With this equation, it is still not possible to express the limit due to the denominator h where if zero is substituted, will be undefined. Therefore, we can check if applying some trigonometric identities can be useful.

Analyzing our equation, we can observe that both the first and second terms in the numerator of the limit is a tangent of a sum of two angles x and h and a tangent of angle x. With this observation, we can try to apply the defining relation identities for tangent, sine, and cosine. Applying this, we have

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\frac{ \frac{\sin{(x+h)}}{\cos{(x+h)}} – \frac{\sin{(x)}}{\cos{(x)}} }{h}}$$

Algebraically re-arranging by applying some rules of fraction, we have

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\frac{ \frac{\sin{(x+h)}\cos{(x)} – \cos{(x+h)}\sin{(x)}}{\cos{(x+h)}\cos{(x)}} }{h}}$$

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\frac{ \sin{(x+h)}\cos{(x)} – \cos{(x+h)}\sin{(x)} }{h\cos{(x+h)}\cos{(x)}}}$$

Looking at the re-arranged numerator, we can try to apply the sum and difference identities for sine and cosine, also called Ptolemy’s identities.

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\frac{ \sin{(x+h-x)} }{h\cos{(x+h)}\cos{(x)}}}$$

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\frac{ \sin{(h)} }{h\cos{(x+h)}\cos{(x)}}}$$

Re-arranging by applying the limit of product of two functions, we have

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\left( \frac{ \sin{(h)} }{h} \cdot \frac{1}{\cos{(x+h)}\cos{(x)}} \right)}$$

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\left( \frac{ \sin{(h)} }{h} \right)} \cdot \lim \limits_{h \to 0} {\left(\frac{1}{\cos{(x+h)}\cos{(x)}} \right)}$$

In accordance with the limits of trigonometric functions, the limit of trigonometric function $latex \sin{(\theta)}$ to $latex \theta$ as $latex \theta$ approaches zero is equal to one. The same can be applied to $latex \sin{(h)}$ over $latex h$. Applying, we have

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {1} \cdot \lim \limits_{h \to 0} {\left(\frac{1}{\cos{(x+h)}\cos{(x)}} \right)}$$

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\left(\frac{1}{\cos{(x+h)}\cos{(x)}} \right)}$$

Finally, we have successfully made it possible to evaluate the limit of whatever is left in the equation. Evaluating by substituting the approaching value of $latex h$, we have

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\left(\frac{1}{\cos{(x+h)}\cos{(x)}} \right)}$$

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\left(\frac{1}{\cos{(x+(0))}\cos{(x)}} \right)}$$

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\left(\frac{1}{\cos{(x)}\cos{(x)}} \right)}$$

## Practice of derivatives of compound tangent functions

Derivatives of tangent quiz
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