Summarize this article:
Last updated on September 22, 2025
We use the derivative of u to understand how the function changes in response to a slight change in x. Derivatives are useful in various real-life applications, such as calculating profit or loss. We will now discuss the derivative of u in detail.
We now understand the derivative of u. It is commonly represented as d/dx (u) or (u)', and its derivative depends on the specific function u represents.
If u is a differentiable function, its derivative will be well-defined within its domain.
The key concepts to keep in mind include the rules of differentiation and how they apply to various forms of functions.
The derivative of u can be denoted as d/dx (u) or (u)'.
The specific formula for differentiating u depends on the form of the function u. For instance, if u = tan x, then d/dx (u) = sec²x.
The formula applies to all x where the function u is defined and differentiable.
We can derive the derivative of u using various proofs depending on the function u. To demonstrate this, we use differentiation rules and identities. Some methods include:
By First Principle - The derivative of a function like u can be shown using the First Principle, which expresses the derivative as the limit of the difference quotient.
Using Chain Rule-The chain rule helps in differentiating composite functions.
Using Product Rule-The product rule is used when u is the product of two or more functions. Here, we will demonstrate the differentiation of specific functions like tan x as an example:
For f(x) = tan x, the derivative can be expressed as the following limit: f'(x) = limₕ→₀ [f(x + h) - f(x)] / h Given f(x) = tan x, we write f(x + h) = tan(x + h).
Substituting these into the equation, we simplify and find f'(x) = sec²x.
For tan x = sin x/cos x, applying the chain and quotient rule gives: d/dx(tan x) = sec²x.
For tan x = sin x · (cos x)⁻¹, we apply the product rule and find the derivative to be sec²x.
When a function is differentiated multiple times, the resulting derivatives are called higher-order derivatives. These can be complex to understand.
Consider a car, where the speed changes (first derivative) and the rate at which the speed changes (second derivative) also changes.
Higher-order derivatives help us understand more about functions like u. For example, if u = tan x, the first derivative is f′(x), and subsequent derivatives can be found similarly.
Special cases arise when the function u has undefined points or behaves irregularly. For instance, if u = tan x, the derivative is undefined at x = π/2 due to a vertical asymptote. At x = 0, the derivative is sec²(0), which equals 1.
Errors frequently occur when differentiating functions like u. These mistakes can be avoided by understanding the correct procedures. Here are a few common mistakes and how to avoid them:
Calculate the derivative of (u·v) where u = tan x and v = sec²x
Here, f(x) = tan x · sec²x. Using the product rule, f'(x) = u′v + uv′. In the given equation, u = tan x and v = sec²x.
Let’s differentiate each term: u′ = d/dx (tan x) = sec²x v′ = d/dx (sec²x) = 2 sec²x tan x Substituting into the equation, f'(x) = sec⁴x + 2 sec²x tan²x.
Thus, the derivative of the specified function is sec⁴x + 2 sec²x tan²x.
Divide the function into two parts. Find each derivative and combine them using the product rule to get the final result.
An engineering firm is designing a slope represented by y = tan(x) where y is the elevation at a distance x. If x = π/4 meters, measure the slope of the slope.
We have y = tan(x) (slope of the slope)...(1)
Differentiate equation (1): dy/dx = sec²(x) Given x = π/4, substitute this into the derivative: sec²(π/4) = 2 (since tan(π/4) = 1)
Hence, the slope at x = π/4 is 2.
The slope at x = π/4 is 2, indicating that at this point, the elevation rises at a rate twice the horizontal distance.
Derive the second derivative of the function y = tan(x).
First, find the first derivative: dy/dx = sec²(x)...(1)
Now differentiate equation (1) for the second derivative: d²y/dx² = d/dx [sec²(x)]
Use the product rule: d²y/dx² = 2 sec(x) · [sec(x) tan(x)] = 2 sec²(x) tan(x)
Therefore, the second derivative of y = tan(x) is 2 sec²(x) tan(x).
Start with the first derivative, then use the product rule to differentiate sec²(x) and simplify to find the second derivative.
Prove: d/dx (tan²(x)) = 2 tan(x) sec²(x).
Using the chain rule, consider y = tan²(x). Rewriting, [tan(x)]².
Differentiate using the chain rule: dy/dx = 2 tan(x) · d/dx [tan(x)]
Since the derivative of tan(x) is sec²(x), dy/dx = 2 tan(x) · sec²(x). Thus, d/dx (tan²(x)) = 2 tan(x) sec²(x).
Use the chain rule to differentiate, replace tan(x) with its derivative, and substitute y = tan²(x) to derive the equation.
Solve: d/dx (u/x) where u = tan x
Differentiate using the quotient rule: d/dx (tan x/x) = (d/dx (tan x) · x - tan x · d/dx(x))/ x²
Substitute d/dx (tan x) = sec²x and d/dx(x) = 1: (x sec²x - tan x) / x² = (x sec²x - tan x) / x²
Therefore, d/dx (tan x/x) = (x sec²x - tan x) / x²
Differentiate the function using the product and quotient rules, then simplify to obtain the final result.
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.
: He loves to play the quiz with kids through algebra to make kids love it.