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Last updated on October 8, 2025
We explore the derivative of tan²(x), which is 2 tan(x) sec²(x), as a tool for understanding how the function tan²(x) changes concerning changes in x. Derivatives are crucial in calculating various changes in real-life situations. Let's delve into the details of the derivative of tan²(x).
The derivative of tan²x can be represented as d/dx (tan²x) or (tan²x)', and its value is 2 tan(x) sec²(x). The function tan²x is differentiable within its domain.
Key concepts include:
Tangent Function: tan(x) = sin(x)/cos(x).
Chain Rule: Used for differentiating composite functions like tan²(x).
Secant Function: sec(x) = 1/cos(x).
The derivative of tan²x is denoted as d/dx (tan²x) or (tan²x)'.
The formula is: d/dx (tan²x) = 2 tan(x) sec²(x)
This formula is applicable for all x where cos(x) ≠ 0.
To derive the derivative of tan²x, we use various methods of differentiation, including:
Let y = tan²(x) = (tan(x))². Differentiating using the chain rule: dy/dx = 2 tan(x) d/dx (tan(x)) Since d/dx (tan(x)) = sec²(x), dy/dx = 2 tan(x) sec²(x). Hence, proved.
Express tan²(x) as (tan(x))(tan(x)) and use the product rule: d/dx (tan²x) = d/dx [(tan(x))(tan(x))] = tan(x) d/dx (tan(x)) + tan(x) d/dx (tan(x)) = tan(x) sec²(x) + tan(x) sec²(x) = 2 tan(x) sec²(x). Hence, proved.
Higher-order derivatives result from differentiating a function multiple times. They provide insights into the function's behavior.
For tan²(x), the first derivative is f′(x), which describes the rate of change, while the second derivative, f′′(x), indicates the rate of that change. This pattern continues for higher derivatives.
When x is π/2, the derivative is undefined due to a vertical asymptote in tan(x).
When x is 0, the derivative of tan²x = 2 tan(0) sec²(0) = 0.
Students often make errors when differentiating tan²x. Understanding the correct methods helps avoid these mistakes.
Calculate the derivative of (tan²x · sec²x).
For f(x) = tan²x · sec²x, use the product rule: f'(x) = u′v + uv′ Where u = tan²x and v = sec²x. u′ = d/dx (tan²x) = 2 tan(x) sec²(x) v′ = d/dx (sec²x) = 2 sec²x tan(x) Substituting: f'(x) = (2 tan(x) sec²(x)) sec²x + tan²x (2 sec²x tan(x)) = 2 tan(x) sec⁴x + 2 tan³x sec²x.
Differentiate the given function using the product rule.
Break it into two parts, find their derivatives, and combine them.
The elevation of a hill is represented by y = tan²(x) where y is the height at a distance x. If x = π/6, find the slope of the hill.
Given y = tan²(x), differentiate: dy/dx = 2 tan(x) sec²(x) Substitute x = π/6: dy/dx = 2 tan(π/6) sec²(π/6) tan(π/6) = 1/√3, sec(π/6) = 2/√3 dy/dx = 2(1/√3)(4/3) = 8/9.
Find the slope by differentiating y = tan²(x) and substituting x = π/6.
Calculate the result using trigonometric values.
Derive the second derivative of the function y = tan²(x).
First derivative: dy/dx = 2 tan(x) sec²(x). Now, differentiate again: d²y/dx² = d/dx [2 tan(x) sec²(x)] Use the product rule: d²y/dx² = 2[sec²(x) sec²(x) + tan(x) 2 sec²(x) tan(x)] = 2 sec⁴(x) + 4 tan²(x) sec²(x).
Use the product rule for the second derivative.
Differentiate the first derivative and simplify.
Prove: d/dx (tan²(x)) = 2 tan(x) sec²(x).
Use the chain rule: y = tan²(x) = (tan(x))² dy/dx = 2 tan(x) d/dx (tan(x)) = 2 tan(x) sec²(x), hence proved.
Differentiate using the chain rule by considering the inner and outer functions, then simplify.
Solve: d/dx (tan²x / x).
Use the quotient rule: d/dx (tan²x / x) = (x d/dx (tan²x) - tan²x d/dx (x)) / x² = (x (2 tan(x) sec²(x)) - tan²x) / x² = (2x tan(x) sec²(x) - tan²x) / x².
Differentiate using the quotient rule.
Simplify the result by finding the derivatives of the numerator and denominator.
Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.
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