Last updated on August 5th, 2025
We use the derivative of 4tan(x), which is 4sec²(x), as a measuring tool for how the function 4tan(x) changes in response to a slight change in x. Derivatives help us calculate profit or loss in real-life situations. We will now talk about the derivative of 4tan(x) in detail.
We now understand the derivative of 4tanx. It is commonly represented as d/dx (4tanx) or (4tanx)', and its value is 4sec²x. The function 4tanx has a clearly defined derivative, indicating it is differentiable within its domain. The key concepts are mentioned below: Tangent Function: (tan(x) = sin(x)/cos(x)). Quotient Rule: Rule for differentiating tan(x) (since it consists of sin(x)/cos(x)). Secant Function: sec(x) = 1/cos(x).
The derivative of 4tanx can be denoted as d/dx (4tanx) or (4tanx)'. The formula we use to differentiate 4tanx is: d/dx (4tanx) = 4sec²x (or) (4tanx)' = 4sec²x The formula applies to all x where cos(x) ≠ 0.
We can derive the derivative of 4tanx using proofs. To show this, we will use the trigonometric identities along with the rules of differentiation. There are several methods we use to prove this, such as: By First Principle Using Chain Rule Using Product Rule We will now demonstrate that the differentiation of 4tanx results in 4sec²x using the above-mentioned methods: By First Principle The derivative of 4tanx can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient. To find the derivative of 4tanx using the first principle, we will consider f(x) = 4tanx. Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1) Given that f(x) = 4tanx, we write f(x + h) = 4tan(x + h). Substituting these into equation (1), f'(x) = limₕ→₀ [4tan(x + h) - 4tanx] / h = 4limₕ→₀ [tan(x + h) - tanx] / h = 4limₕ→₀ [ [sin(x + h) / cos(x + h)] - [sinx / cosx] ] / h = 4limₕ→₀ [ [sin(x + h) cosx - cos(x + h) sinx] / [cosx · cos(x + h)] ] / h We now use the formula sinA cosB - cosA sinB = sin(A - B). f'(x) = 4limₕ→₀ [ sinh ] / [ h cosx · cos(x + h)] = 4limₕ→₀ (sinh)/h · limₕ→₀ 1 / [cosx · cos(x + h)] Using limit formulas, limₕ→₀ (sinh)/h = 1. f'(x) = 4 [ 1 / (cosx · cos(x + 0))] = 4/cos²x As the reciprocal of cosine is secant, we have, f'(x) = 4sec²x. Hence, proved. Using Chain Rule To prove the differentiation of 4tanx using the chain rule, We use the formula: 4tanx = 4(sinx/cosx) Consider f(x) = sinx and g(x) = cosx So we get, 4tanx = 4f(x)/g(x) By quotient rule: d/dx [f(x) / g(x)] = [f '(x) g(x) - f(x) g'(x)] / [g(x)]²… (1) Let’s substitute f(x) = sinx and g(x) = cosx in equation (1), d/dx (4tanx) = 4[(cosx)(cosx)- (sinx)(-sinx)] / (cosx)² (cos²x + sin²x)/ cos²x …(2) Here, we use the formula: (cos²x) + (sin²x) = 1 (Pythagorean identity) Substituting this into (2), d/dx (4tanx) = 4/ (cosx)² Since secx = 1/cosx, we write: d/dx(4tanx) = 4sec²x Using Product Rule We will now prove the derivative of 4tanx using the product rule. The step-by-step process is demonstrated below: Here, we use the formula, 4tanx = 4(sinx/cosx) 4tanx = 4(sinx)(cosx)⁻¹ Given that, u = 4sinx and v = (cosx)⁻¹ Using the product rule formula: d/dx [u.v] = u'v + uv' u' = d/dx (4sinx) = 4cosx. (substitute u = 4sinx) Here we use the chain rule: v = (cosx)⁻¹ = (cosx)⁻¹ (substitute v = (cosx)⁻¹) v' = -1(cos)⁻². d/dx (cosx) v' = sinx / (cosx)² Again, use the product rule formula: d/dx (4tanx) = u'v + uv' Let’s substitute u = 4sinx, u' = 4cosx, v = (cosx)⁻¹, and v' = sinx / (cosx)² When we simplify each term: We get, d/dx (4tanx) = 4 + 4sin²x / (cosx)² sin²x / (cosx)² = tan²x (we use the identity sin²x + cos²x = 1) Thus: d/dx (4tanx) = 4 + 4tan²x Since, 4 + 4tan²x = 4sec²x d/dx (4tanx) = 4sec²x.
When a function is differentiated several times, the derivatives obtained are referred to as higher-order derivatives. Higher-order derivatives can be a little tricky. To understand them better, think of a car where the speed changes (first derivative) and the rate at which the speed changes (second derivative) also changes. Higher-order derivatives make it easier to understand functions like 4tan(x). For the first derivative of a function, we write f′(x), which indicates how the function changes or its slope at a certain point. The second derivative is derived from the first derivative, which is denoted using f′′(x). Similarly, the third derivative, f′′′(x), is the result of the second derivative, and this pattern continues. For the nth Derivative of 4tan(x), we generally use fⁿ(x) for the nth derivative of a function f(x), which tells us the change in the rate of change (continuing for higher-order derivatives).
When x is π/2, the derivative is undefined because tan(x) has a vertical asymptote there. When x is 0, the derivative of 4tanx = 4sec²(0), which is 4.
Students frequently make mistakes when differentiating 4tanx. These mistakes can be resolved by understanding the proper solutions. Here are a few common mistakes and ways to solve them:
Calculate the derivative of (4tanx·sec²x)
Here, we have f(x) = 4tanx·sec²x. Using the product rule, f'(x) = u'v + uv' In the given equation, u = 4tanx and v = sec²x. Let’s differentiate each term, u' = d/dx (4tanx) = 4sec²x v' = d/dx (sec²x) = 2sec²x tanx Substituting into the given equation, f'(x) = (4sec²x)(sec²x) + (4tanx)(2sec²x tanx) Let’s simplify terms to get the final answer, f'(x) = 4sec⁴x + 8sec²x tan²x Thus, the derivative of the specified function is 4sec⁴x + 8sec²x tan²x.
We find the derivative of the given function by dividing the function into two parts. The first step is finding its derivative and then combining them using the product rule to get the final result.
A bridge is designed where its elevation is represented by the function y = 4tan(x) where y represents the height of the bridge at a distance x. If x = π/6 meters, measure the slope of the bridge.
We have y = 4tan(x) (slope of the bridge)...(1) Now, we will differentiate the equation (1) Take the derivative of 4tan(x): dy/dx = 4sec²(x) Given x = π/6 (substitute this into the derivative) sec²(π/6) = 1 + tan²(π/6) sec²(π/6) = 1 + (1/√3)² = 4/3 (since tan(π/6) = 1/√3) Hence, we get the slope of the bridge at a distance x = π/6 as 16/3.
We find the slope of the bridge at x = π/6 as 16/3, which means that at a given point, the height of the bridge would rise at a rate of 16/3 times the horizontal distance.
Derive the second derivative of the function y = 4tan(x).
The first step is to find the first derivative, dy/dx = 4sec²(x)...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [4sec²(x)] Here we use the product rule, d²y/dx² = 8sec(x)·d/dx [sec(x)] d²y/dx² = 8sec(x)·[sec(x) tan(x)] 8sec²(x) tan(x) Therefore, the second derivative of the function y = 4tan(x) is 8sec²(x) tan(x).
We use the step-by-step process, where we start with the first derivative. Using the product rule, we differentiate sec²(x). We then substitute the identity and simplify the terms to find the final answer.
Prove: d/dx (4tan²(x)) = 8tan(x) sec²(x).
Let’s start using the chain rule: Consider y = 4tan²(x) = 4[tan(x)]² To differentiate, we use the chain rule: dy/dx = 8tan(x)·d/dx [tan(x)] Since the derivative of tan(x) is sec²(x), dy/dx = 8tan(x)·sec²(x) Substituting y = 4tan²(x), d/dx (4tan²(x)) = 8tan(x)·sec²(x) Hence proved.
In this step-by-step process, we used the chain rule to differentiate the equation. Then, we replace tan(x) with its derivative. As a final step, we substitute y = 4tan²(x) to derive the equation.
Solve: d/dx (4tanx/x)
To differentiate the function, we use the quotient rule: d/dx (4tanx/x) = (d/dx (4tanx)·x - 4tanx·d/dx(x))/x² We will substitute d/dx (4tanx) = 4sec²x and d/dx(x) = 1 (4sec²x·x - 4tanx·1) / x² = (4x sec²x - 4tanx) / x² = 4x sec²x - 4tanx / x² Therefore, d/dx (4tanx/x) = 4x sec²x - 4tanx / x²
In this process, we differentiate the given function using the quotient rule. As a final step, we simplify the equation to obtain the final result.
Derivative: The derivative of a function indicates how the given function changes in response to a slight change in x. Tangent Function: The tangent function is one of the primary six trigonometric functions and is written as tan x. Secant Function: A trigonometric function that is the reciprocal of the cosine function. It is typically represented as sec x. First Derivative: It is the initial result of a function, which gives us the rate of change of a specific function. Asymptote: The function goes near a line without intersecting or crossing it. This line is known as an asymptote.
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