# 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

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

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

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

$latex \frac{d}{dx} f(x) = \frac{1}{\cos{(x)}\cos{(x)}}$$We know that by the defining relation identities, the reciprocal of the trigonometric function cosine is secant. Applying, we have$$\frac{d}{dx} f(x) = \frac{1}{\cos{(x)}} \cdot \frac{1}{\cos{(x)}}\frac{d}{dx} f(x) = \sec{(x)} \cdot \sec{(x)}\frac{d}{dx} f(x) = \sec^{2}{(x)}$$Therefore, the derivative of the trigonometric function ‘tangent‘ is:$$\frac{d}{dx} (\tan{(x)}) = \sec^{2}{(x)}$$## How to derive a Tangent Function? The derivative process of a tangent function is very straightforward assuming you have already learned the concepts behind the usage of the tangent function and how we arrived to its derivative formula. ### METHOD 1: Derivative of Tangent of any angle x in terms of the same angle x. Step 1: Analyze if the tangent of an angle is a function of that same angle. For example, if the right-hand side of the equation is$latex \tan{(x)}$, then check if it is a function of the same angle x or f(x). After this, proceed to Step 2 until you complete the derivation steps. Note: If$latex \tan{(x)}$is a function of a different angle or variable such as f(t) or f(y), it will use implicit differentiation which is out of the scope of this article. Step 2: Then directly apply the derivative formula of the tangent function$latex \frac{dy}{dx} = \sec^{2}{(x)}$If nothing is to be simplified anymore, then that would be the final answer. ### METHOD 2: Derivative of Tangent of any function u in terms of x. Step 1: Express the function as$latex F(x) = \tan{(u)}$, where$latex u$represents any function other than x. Step 2: Consider$latex \tan{(u)}$as the outside function$latex f(u)$and$latex u$as the inner function$latex g(x)$of the composite function$latex F(x)$. Hence we have$latex f(u) = \tan{(u)}$and$latex g(x) = u$Step 3: Get the derivative of the outer function$latex f(u)$, which must use the derivative of the tangent function, in terms of$latex u$.$latex \frac{d}{du} \left( \tan{(u)} \right) = \sec^{2}{(u)}$Step 4: Get the derivative of the inner function$latex g(x) = u$. Use the appropriate derivative rule that applies to$latex u$. Step 5: Apply the basic chain rule formula by algebraically multiplying the derivative of outer function$latex f(u)$by the derivative of inner function$latex g(x)latex \frac{dy}{dx} = \frac{d}{du} (f(u)) \cdot \frac{d}{dx} (g(x))latex \frac{dy}{dx} = \sec^{2}{(u)} \cdot \frac{d}{dx} (u)$Step 6: Substitute$latex u$into$latex f'(u)$Step 7: Simplify and apply any function law whenever applicable to finalize the answer. ## Graph of Tangent of x VS. The Derivative of Tangent of x Given the function$latex f(x) = \tan{(x)}$the graph is illustrated as And as we know by now, by deriving$latex f(x) = \tan{(x)}$, we get$latex f'(x) = \sec^{2}{(x)}$which is illustrated graphically as Illustrating both graphs in one, we have Analyzing the differences of these functions through these graphs, you can observe that the original function$latex f(x) = \tan{(x)}$has a domain of$latex \left(-\frac{3\pi}{2},-\frac{\pi}{2}\right) \cup \left(-\frac{\pi}{2},\frac{\pi}{2}\right) \cup \left(\frac{\pi}{2},\frac{3\pi}{2}\right)$within the finite intervals of$latex \left(-\frac{3\pi}{2},\frac{3\pi}{2}\right)$and exists within the range of$latex (-\infty,\infty)$or all real numbers whereas the derivative$latex f'(x) = \sec^{2}{(x)}$has a domain of$latex \left(-\frac{3\pi}{2},-\frac{\pi}{2}\right) \cup \left(-\frac{\pi}{2},\frac{\pi}{2}\right) \cup \left(\frac{\pi}{2},\frac{3\pi}{2}\right)$within the finite intervals of$latex \left(-\frac{3\pi}{2},\frac{3\pi}{2}\right)$and exists within the range of$latex [1,\infty)$or$latex y \geq 1$## Examples Below are some examples of using either the first or second method in deriving a tangent function. ### EXAMPLE 1 Derive:$latex f(\beta) = \tan{(\beta)}$Solution: Analyzing the given tangent function, we see that it is only a tangent of a single angle$latex \beta$. Therefore, we can use the first method to derive this problem. Step 1: Analyze if the tangent of$latex \beta$is a function of$latex \beta$. In this problem, it is. Hence, proceed to step 2. Step 1: Directly apply the derivative formula of the tangent function and derive in terms of$latex \beta$. Since no further simplification is needed, the final answer is:$latex f'(\beta) = \sec^{2}{(\beta)}$### EXAMPLE 2 Derive:$latex F(x) = \tan{\left(3x^2+6 \right)}$Solution: Analyzing the given tangent function, we see that it is a tangent of a polynomial function. Therefore, we can use the second method to derive this problem. Step 1: Express the tangent function as$latex F(x) = \tan{(u)}$, where$latex u$represents any function other than x. In this problem,$latex u = 3x^2+6$We will substitute this later as we finalize the derivative of the problem. Step 2: Consider$latex \tan{(u)}$as the outside function$latex f(u)$and$latex u$as the inner function$latex g(x)$of the composite function$latex F(x)$. For this problem, we have$latex f(u) = \tan{(u)}$and$latex g(x) = u = 3x^2+6$Step 3: Get the derivative of the outer function$latex f(u)$, which must use the derivative of the tangent function, in terms of$latex u$.$latex \frac{d}{du} \left( \tan{(u)} \right) = \sec^{2}{(u)}$Step 4: Get the derivative of the inner function$latex g(x)$or$latex u$. Since our$latex u$in this problem is a polynomial function, we will use power rule and sum/difference of derivatives to derive$latex u$.$latex \frac{d}{dx}(g(x)) = \frac{d}{dx} \left(3x^2+6 \right)latex \frac{d}{dx}(g(x)) = 6x$Step 5: Apply the basic chain rule formula by algebraically multiplying the derivative of outer function$latex f(u)$by the derivative of inner function$latex g(x)latex \frac{dy}{dx} = \frac{d}{du} (f(u)) \cdot \frac{d}{dx} (g(x))latex \frac{dy}{dx} = \sec^{2}{(u)} \cdot 6x$Step 6: Substitute$latex u$into$latex f'(u)latex \frac{dy}{dx} = \sec^{2}{(u)} \cdot 6xlatex \frac{dy}{dx} = \sec^{2}{(3x^2+6)} \cdot 6x$Step 7: Simplify and apply any function law whenever applicable to finalize the answer.$latex \frac{dy}{dx} = \sec^{2}{(3x^2+6)} \cdot 6xlatex \frac{dy}{dx} = 6x\sec^{2}{(3x^2+6)}$And the final answer is:$latex F'(x) = 6x\sec^{2}{(3x^2+6)}$or$latex F'(x) = 6x\sec^{2}{(3(x^2+2))}\$  