Derivative of sin 5x
2026-02-28 23:18 Diff
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Last updated on September 27, 2025

The derivative of sin(5x) helps us understand how the sine function changes in response to a slight change in x. Derivatives are useful in various real-life applications, including calculating profit or loss. We will now explore the derivative of sin(5x) in detail.

What is the Derivative of sin 5x?

The derivative of sin 5x is denoted as d/dx (sin 5x) or (sin 5x)', and its value is 5cos(5x). This indicates that the function sin 5x is differentiable within its domain.

Key concepts include: 

Sine Function: sin(5x) represents the sine function with an argument of 5x. 

Chain Rule: Used for differentiating composite functions like sin(5x). 

Cosine Function: cos(x) is the derivative of sin(x), and it appears in the derivative formula for sin(5x).

Derivative of sin 5x Formula

The derivative of sin 5x can be represented as d/dx (sin 5x) or (sin 5x)'. Using the chain rule, the formula is: d/dx (sin 5x) = 5 cos(5x)

This formula applies for all x, given that sin 5x is a continuous function.

Proofs of the Derivative of sin 5x

We can derive the derivative of sin 5x using different methods.

These include: 

  • By First Principle 
     
  • Using the Chain Rule

By First Principle

To prove by the first principle, consider f(x) = sin 5x. Its derivative is expressed as the limit of the difference quotient: f'(x) = limₕ→₀ [f(x + h) - f(x)] / h Substituting f(x) = sin 5x, we have: f'(x) = limₕ→₀ [sin(5(x + h)) - sin(5x)] / h Applying the sine difference identity and limit laws, the result is: f'(x) = 5 cos(5x)

Using the Chain Rule

Let u = 5x, so that f(x) = sin(u). Then, f'(x) = cos(u) * du/dx. Since du/dx = 5, we have: d/dx (sin 5x) = cos(5x) * 5 = 5 cos(5x)

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Higher-Order Derivatives of sin 5x

Higher-order derivatives are obtained by repeatedly differentiating a function. For sin 5x, the first derivative is 5 cos(5x), the second derivative is -25 sin(5x), and so on.

 These derivatives help us understand changes at multiple levels, much like how acceleration (second derivative) relates to speed (first derivative).

Special Cases:

When x is 0, the derivative of sin 5x = 5 cos(0) = 5. 

At multiples of π/5, the derivative equals 0 because cos(nπ) = 0 for n = ±1, ±2, ...

Common Mistakes and How to Avoid Them in Derivatives of sin 5x

When differentiating sin 5x, students often make errors. Understanding the correct approach can help. Here are common mistakes and solutions:

Problem 1

Calculate the derivative of sin 5x·cos 5x

Okay, lets begin

Let f(x) = sin 5x · cos 5x. Using the product rule, f'(x) = u'v + uv'. Here, u = sin 5x and v = cos 5x. Differentiating each, u' = 5 cos 5x and v' = -5 sin 5x. Substituting, we get f'(x) = (5 cos 5x)(cos 5x) + (sin 5x)(-5 sin 5x) = 5 cos²5x - 5 sin²5x.

Explanation

We applied the product rule by differentiating each part and combining the results.

Simplifying gives the final answer.

Well explained 👍

Problem 2

A pendulum swings according to the function y = sin(5x), where y is the displacement at time x. Find the rate of change of displacement when x = π/10.

Okay, lets begin

The function is y = sin(5x). Differentiate to get dy/dx = 5 cos(5x). At x = π/10, dy/dx = 5 cos(5(π/10)) = 5 cos(π/2) = 0. The rate of change of displacement is 0 at x = π/10, indicating a momentary pause in motion.

Explanation

At x = π/10, the pendulum reaches a peak or trough, so the rate of change is zero, reflecting no motion at that instant.

Well explained 👍

Problem 3

Find the second derivative of y = sin 5x.

Okay, lets begin

First, find dy/dx = 5 cos 5x. Then, the second derivative is d²y/dx² = d/dx(5 cos 5x) = -25 sin 5x.

Explanation

We first differentiated to find the first derivative, then differentiated again to find the second derivative, applying chain rules where necessary.

Well explained 👍

Problem 4

Prove: d/dx (sin²(5x)) = 10 sin(5x) cos(5x).

Okay, lets begin

Let y = sin²(5x). Using the chain rule, dy/dx = 2 sin(5x) d/dx(sin(5x)) = 2 sin(5x) (5 cos(5x)) = 10 sin(5x) cos(5x).

Explanation

The chain rule was applied first to the outer function, then to the inner function, and results were multiplied together.

Well explained 👍

Problem 5

Differentiate (sin 5x)/x.

Okay, lets begin

Using the quotient rule: d/dx [(sin 5x)/x] = [x(5 cos 5x) - sin 5x(1)] / x² = (5x cos 5x - sin 5x) / x².

Explanation

We used the quotient rule by identifying the numerator and denominator functions, differentiating each, and simplifying the result.

Well explained 👍

FAQs on the Derivative of sin 5x

1.What is the derivative of sin 5x?

The derivative of sin 5x is 5 cos(5x), using the chain rule.

2.How is the derivative of sin 5x applied in real life?

It can be used to model oscillations or wave behaviors, such as in physics for pendulums or signals.

3.Is the derivative of sin 5x defined for all x?

Yes, sin 5x is differentiable for all x, as it is a continuous function.

4.How can the chain rule help with differentiating sin 5x?

The chain rule helps because sin 5x is a composite function, requiring differentiation of the outer sin function and the inner 5x function.

5.What is the second derivative of sin 5x?

The second derivative of sin 5x is -25 sin 5x, obtained by differentiating 5 cos 5x.

Important Glossaries for the Derivative of sin 5x

  • Derivative: Describes how a function changes as its input changes.
  • Sine Function: Denoted as sin(x), a basic trigonometric function.
  • Cosine Function: Denoted as cos(x), derivative of sin(x).
  • Chain Rule: A fundamental rule for differentiating composite functions.
  • Quotient Rule: Used for differentiating ratios of functions.

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Jaskaran Singh Saluja

About the Author

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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