So, once you are in high school, trigonometry and calculus become two key aspects of math. You cannot almost do anything without either one of them or both. So, you just do not need them for solving sums in math. They are there in physics, chemistry, economics, statistics- everything, in short. One of the most important reasons behind these is their importance in the formation of curves. So, curves are fundamental to every scientific study as you engage in higher study. Therefore, both trigonometry and calculus help you find them and explain them. Now, differentiation is breaking an amount into negligibly small quantities. On the other hand, integration is the adding up of these extremely small quantities to form a single entity. However, it is also a measure of the rate of change of any quantity, such as the derivative of tangent, for example.

Now, we already know that the basic trigonometric functions are sin, cos, tan, cosec, sec, and cot. So, tan stands for tangent. Now, amongst all functions, whenever the question of derivative comes, the derivative is another trigonometric function itself. Like, it is pretty obvious that the derivative of tangent cannot be x^2, for example. So, there are certain rules that you need to remember while applying calculus to trigonometric values. In this article, we will concentrate solely on the derivative of tangent and go through its value, examples, and so on.

**Derivative of tangent functions**

Now, tangent is the ratio of the perpendicular to the base of the triangle. On the other hand, we know that cosine is the ratio of the base to the hypotenuse of the triangle. Therefore, sec is the ratio of the hypotenuse to the base of the triangle. So now, the derivative of tangent is a function of the sec of the said angle. Let us see how.

So, let us consider the angle with respect to which we will calculate the trigonometric functions to be x.

Therefore, we have sin x, cos x, and tan x. Now, from trigonometry, we already know that tan x = sin x/ cos x. So this is because sin x is the ratio of the perpendicular to the hypotenuse of the triangle. Therefore, if you divide both, the hypotenuse gets canceled off. Moreover, we already know that sec x = 1/ cos^2x.

Now, let us see the derivative of tangent.

So, we will be applying the quotient rule here. According to the quotient rule, you have to differentiate the numerator and multiply it with the denominator. Then, you have to differentiate the denominator and multiply it with the numerator and subtract it from the first term. So finally you have to divide the entire expression by the square of the denominator.

**Derivative of tangent proof**

Therefore, as per this-

(tan x)’ = (sin x / cos x)’ = ((sin x)’ * cos x – (cos x)’ * sin x) / cos^2x

Now, we all know that (sin x)’ = cos x. Moreover, (cos x)’ = – sin x.

Or, (tan x)’ = (cos^2 x + sin^2 x) / cos^2 x

Now, we already know cos^2 x + sin^2 x = 1.

So, (tan x)’ = 1/ cos^2 x

Or, (tan x)’ = sec^2 x.

Therefore, the derivative of tangent is sec^2 x.

**Derivative of tangent inverse**

So, in trigonometry, we get inverse functions as well. For example, the inverse of tangent is written as tan^(-1) x. The derivative of tangent inverse is not the same as the normal derivative of tangent. So, it is 1/ (1 + x^2)

Therefore, let us see how.

So, we have the function f(x) = tan^(-1)x

Hence, let y = tan^(-1)x

So, tan y = x (This is because tan(tan^(-1)x) = x)

⇒ x = tan y → (1)

Now, we will have to differentiate both sides of x = tan y w.r.t. x

(dx/dx) = d(tan y)/dx

So, 1 = [d(tan y)/dx] × [dy/dy] [Multiplying and dividing by ‘dy’]

Or, 1 = [d(tan y)/dy] × [dy/dx]

So, 1 = [sec^2y] dy/dx (we already know that d(tan x)/dx = sec^2x)

Or, 1 = [1 + tan^2y] dy/dx [Using trigonometric laws, sec^2y = 1 + tan^2y]

dy/dx = [1] / [1 + tan^2y]

Substituting tan y = x (from (1)) into dy/dx = [1] / [1 + tan^2y], we have

dy/dx = 1 / (1 + x^2)

⇒ d(tan^(-1)x)/dx = 1 / (1 + x^2)

Hence, we have derived the derivative of tan inverse x using implicit differentiation.

**Derivative of tangent squared**

So, we already know the derivative of tangent. Moreover, we have also seen the proof. Therefore, from here, it is not difficult to find the derivative of tangent squared. So, in this case, we will be using the chain rule of differentiation.

Hence, let us say, f(x) = tan x

So, we have already seen that f’(x) = (tan x)’ = sec^2 x

So now, f(x) = tan^2 x

Therefore, f’(x) = (tan^2x)’ = 2tan x * (dy/ dx(tan x))

So, f’(x) = 2 tan x sec^2 x.

Therefore, the derivative of tangent squared are 2 tan x sec^2 x

**Derivative of tangent lines**

So, we already know that drawing curves of trigonometric functions is very important for better understanding and application. Therefore, similarly, we would need derivative of tangents to get the slope that would help us draw the line.

Read Also:Adjacent angle — all the explanations

So, let us say we have to find the equation of the line tangent to the line f(x) = x^2 at the point where x = 2. It is always better if you have the graph next to you. However, we don’t have one right here. You can do without it.

So, the equation of the line given here is f(x) = x^2

Therefore, following the power rule of differentiation, we get

f’(x) = 2x.

Hence, at the point x = 2,

f’(x) = f’(2) = 2x = 2 x 2 = 4.

Therefore, the slope of the line is 4. We can say this because we know that the first derivative of anything is its rate of change which is exactly the definition of the slope. So, the slope of the tangent is the derivative of the original line.

Now, let us get back to the original function f(x) = x^2

So, here the question says that at x = 2, f(x) = x^2.

Therefore, it must be f(2) = 2^2 = 4

Now, we know y is the dependable axis that is dependent on the value of x. Therefore, f(x) is y which is a function of x.

So, when x= 2, y or f(x) = 4.

Hence, we already know 2 points and the slope. Let us now find the equation of the tangent.

y – y1 = m (x – x1)

Or, y – 4 = 4(x -2)

So, y – 4 = 4x – 8

Or, y = 4x – 4

So, this is the equation of the tangent line.

**Note**

However, keep in mind that in this case we have worked with a tangent line and found the derivative of tangent that way. So, it is not the same as the trigonometric derivative of tangent that we had already worked on within the previous section. Here, we are simply considering a tangent or perpendicular line and finding out its equation.

**Derivative of tangent line calculator**

Now, we have already seen how to find the equation of a tangent using derivatives. However, the internet makes things a lot easier. So, these days you have a number of calculators to help you out. The tangent line calculators help you to calculate the values from complex equations. On the other hand, you can simply use it as a tool for verifying your answers. This is because when we calculate manually, there is never a chance that we might not make a mistake. Moreover, they show the entire process stepwise. So, if you have even made a mistake, you can easily spot where you went wrong. Here we will consider one such. So, let us see the steps to use one such calculator.

**Step 1 to use a tangent line calculator**

Hence, in this case, we are using the tangent line calculator from Symbolabs. Click on Symbolabs and you will directly land on their page. So, on opening the calculator you will find a number of sections. In the first part there is a box that looks like a search engine where you have to enter the data. In the section below, you will find the working out of the sum. However, before you enter any data, you will find that the calculator itself suggests a number of examples there for your better understanding.

**Step 2 to use a tangent line calculator**

Now, enter your data in the search engine-like box. Let us say, you enter the “tangent of f(x) = 1/ x^2, (-1, 1)”. So it means at the point (-1, 1). So after this, you will find a red button right beside the search box that reads “go”. Click it for the sum to run on the calculator.

**Step 3 to use a tangent line calculator**

So, it will take a few seconds to run the sum. Once that is done, you will find the full working out of the question step by step in the next section.

Therefore, in this case-

First we will see the derivative of the given line that would give the slope of the tangent. So, f(x) = 1/ x^2

So, f’(x) = (x^-2)’ = – 2x^-3 = – 2/ x^3

Now, we have the point (-1, 1)

So, at (-1, 1) m = – 2/ (-1)^3 = 2.

Now, we will consider the 2-point form to form the equation of the tangent.

Therefore,

y – y1 = m(x – x1)

Or, y – 1 = 2(x – (-1))

So, y – 1 = 2x + 2

Or, y = 2x + 3.

So, here we do not exactly need the derivative of tangent but this is how we get the equation of the tangent line. Again, this is a little different from what we have seen in the previous few sections involving derivative of tangents and their proof.

**Step 4 to use a tangent line calculator**

After the solution, you will find another box in the shape of a search engine. So, in case you are not satisfied with the answer or your own answer does not match, you can type your answer there. Beside it is a red button that reads “verify”. So, click it to run your program.

**Derivative of tangent FAQs**

**How do you take the derivative of tangent?**

Well, to find the derivative of tangent, you can simply use the formula that (tan x)’ = sec^2 x. However, if you want to know how that happens or the proof behind it, scroll up. Under the section on derivative of tangent proof, you will find a detailed working out of how this result comes. It is not arbitrary. Moreover, it is advisable that you check out the proof for better understanding or you will end up mugging without understanding anything and would not be able to apply it.

**What is the derivative of cot?**

The derivative of cot x is -cosec^2x. So, this is because cot x = cos x/ sin x. Now let us differentiate this by applying the quotient rule. Let us consider the function f(x).

So, f’(x) = (cos x/ sin x)’ = ((cos x)’ * sin x – (sin x)’ * cos x)/ sin^2 x

Or, f’(x) = (- sin^2 x – cos^2 x)/ sin^2 x

So, this is because (sin x)’ = cos x while (cos x)’ = – sin x

So, f’(x) = – (sin^2 x + cos^2 x)/ sin^2 x

However, as per the laws of trigonometric identity, we know that sin^2 x + cos^2 x = 1

Hence, f’(x) = – 1/ sin^2 x.

Now, 1/ sin x = cosec x. So, – 1/ sin^2 x = – cosec^2x. Therefore, the derivative of cotangent is – cosec^2 x.

**What is the derivative of tan 2x?**

So, in this case of derivative of tangents, the angle is 2x. Therefore, the derivative would be 2sec^2 (2x) if you follow the rules of differentiation properly. Following the chain rule of differentiation, you need to find the derivative of 2x as well in this case.

**What is the inverse derivative of tan x?**

We have already seen that the derivative of tangent inverse or tan^(-1)x is 1/ (1 + x^2). If you want to know how to check the section of derivative of tangent inverse in detail.

**Is tan^(-1) the same as cot?**

So no, the inverse of tangent is not the same as cotangent. On the other hand, cotangent is the reciprocal of tangent. So, cot x = 1/ tan x and not equals to tan^(-1) x. Therefore, you also understand that 1/ tan x and tan^(-1) x are not the same. So, just because it has an inverse power, you cannot place it in the denominator like normal exponents. You have to remember that this is a trigonometric function. Moreover, we already know trigonometric functions have special conditions and requirements.