Derivative Of Tanx

Unveiling the Derivative of tan x: A Comprehensive Guide

The trigonometric function tangent (tan x) plays a crucial role in various fields, from calculus and physics to engineering and computer graphics. Understanding its derivative is essential for solving numerous problems involving rates of change, optimization, and curve analysis. This article aims to provide a comprehensive exploration of the derivative of tan x, explaining the underlying principles and demonstrating its application through practical examples.

1. Defining the Tangent Function

Before diving into the derivative, let's revisit the definition of tan x. It's defined as the ratio of the sine function to the cosine function:

tan x = sin x / cos x

This definition is crucial because it allows us to leverage the known derivatives of sin x and cos x to find the derivative of tan x.

2. Applying the Quotient Rule

To find the derivative of tan x, we employ the quotient rule, a fundamental tool in differential calculus. The quotient rule states that if we have a function f(x) = g(x) / h(x), then its derivative is:

f'(x) = [h(x)g'(x) - g(x)h'(x)] / [h(x)]²

In our case, g(x) = sin x and h(x) = cos x. We know that the derivative of sin x is cos x (sin'x = cos x) and the derivative of cos x is -sin x (cos'x = -sin x). Substituting these into the quotient rule, we get:

tan'(x) = [cos x cos x - sin x (-sin x)] / (cos x)²
= [cos²x + sin²x] / cos²x

3. Leveraging the Pythagorean Identity

Notice the numerator: cos²x + sin²x. This is a fundamental trigonometric identity, always equaling 1. Therefore, we can simplify the expression:

tan'(x) = 1 / cos²x

4. Introducing the Secant Function

The reciprocal of the cosine function is the secant function (sec x). Therefore, we can express the derivative of tan x more concisely as:

tan'(x) = sec²x

This is the final and most commonly used form of the derivative of tan x.

5. Practical Applications and Examples

The derivative of tan x, sec²x, finds numerous applications in various fields. Let's consider a couple of examples:

Example 1: Finding the slope of a tangent line.

Suppose we have the function y = tan(2x). To find the slope of the tangent line at x = π/8, we first find the derivative:

y' = 2sec²(2x) (using the chain rule)

Then we substitute x = π/8:

y'(π/8) = 2sec²(π/4) = 2(√2)² = 4

Therefore, the slope of the tangent line to y = tan(2x) at x = π/8 is 4.

Example 2: Solving a related rates problem.

Imagine a ladder leaning against a wall. The angle the ladder makes with the ground is changing at a rate of 0.1 radians per second. How fast is the height of the ladder on the wall changing when the angle is π/4 radians?

Let θ be the angle. The height (h) is given by h = L tan θ, where L is the length of the ladder. Differentiating with respect to time (t), we get:

dh/dt = L sec²(θ) dθ/dt

Substituting dθ/dt = 0.1 radians/second and θ = π/4, we can calculate dh/dt, the rate at which the height is changing.

6. Conclusion

This article has demonstrated the derivation of the derivative of tan x, illustrating its derivation using the quotient rule and trigonometric identities. The resulting formula, tan'(x) = sec²x, is a fundamental result in calculus with widespread applications in various fields. Understanding this derivative is essential for anyone working with trigonometric functions and their applications in problem-solving.

Frequently Asked Questions (FAQs)

1. Why is the derivative of tan x always positive? The secant squared of any angle is always positive or zero (it's undefined only when cos x = 0). Therefore, the derivative of tan x is always non-negative.

2. How does the chain rule apply to the derivative of tan(f(x))? The chain rule states that the derivative of tan(f(x)) is sec²(f(x)) f'(x).

3. What is the second derivative of tan x? The second derivative is found by differentiating sec²x, which requires the chain rule and results in 2sec²(x)tan(x).

4. Can the derivative of tan x be expressed in terms of sin x and cos x? Yes, it can be expressed as (1/cos²x) or (1 + tan²x).

5. What are some real-world applications beyond the examples provided? The derivative of tan x is used in calculating the rate of change of angles in projectile motion, analyzing oscillations in physics, and modeling curves in computer graphics.

Search Results:

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Derivative of tanx? - Socratic 2 May 2018 · \frac{1}{cos^2(x)} You may also consider tan(x) as a generic function defined by ratio: f(x) = \frac{f(x)}{g(x)} and apply the rule f'(x) = \frac{f'(x)g(x)-f(x)g'(x ...

Find the derivative of (x+cosx)/tanx? - Socratic 7 Feb 2018 · Derivative is (tanx-sinxtanx+xsec^2x+secx)/tan^2x We use quotient rule, which says that derivative of (f(x))/(g(x)) is (g(x)f'(x)-f(x)g'(x))/[g(x)]^2 Hence derivative ...

How do I differentiate tan(x) - MyTutor Note: Here, we use d/dx f(x) to mean "the derivative of f(x) with respect to x". 1) rewrite tan(x) as sin(x)/cos(x) 2) Apply the quotient rule (or, alternatively, you could use the product rule using …

What is the derivative of #f(x) = tan(x)cos(x)#? - Socratic 2 Nov 2015 · f'(x) = cosx This is a trick or trap question. The long and tedious way to this in by the product rule, then simplify. The quicker way is to observe that f(x) = tanx cosx = sinx/cosx cosx …

Prove that the derivative of tan(x) is sec^2(x). - MyTutor Let y = tan(x) Recall the definition of tan(x) as sin(x)/cos(x) Therefore y = sin(x)/cos(x) Use the quotient rule, which states that for y = f(x)/g(x), dy/dx = (f...

What is the derivative of #-cosx*ln(secx+tanx)#? - Socratic How do you find the 50th derivative of #y=cos(x)# ? How do you find the derivative of #y=cos(x^2)# ? See all questions in Derivative Rules for y=cos(x) and y=tan(x)

What is the derivative of 5^tanx? - Socratic 3 Mar 2018 · We are trying to find #(dy)/(dx) # when #y = 5^tanx #. Taking natural logs on both sides... #=> ln y = ln 5^tanx #

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How do you find the derivative of 5^tanx? - Socratic 30 Nov 2016 · The formula to differentiate is. #d/dx(a^u)=a^u*ln a*d/dx(u)# therefore. #d/dx(5^tan x)=5^tanx*ln 5*d/dx(tan x)# #d/dx(5^tan x)=5^tanx*ln 5*sec^2 x*(dx/dx)#