The n-th derivative of sinx and cosx with respect to x are equal to sin(nπ/2 +x) and cos(nπ/2 +x) respectively. In this article, let us learn how to differentiate sinx and cosx with respect to x n-times.

The n^{th} order derivative of sinx and cosx are respectively denoted by $\dfrac{d^n}{dx^n}$(sinx) and $\dfrac{d^n}{dx^n}$(cosx). So their formulas are given as follows:

- $\dfrac{d^n}{dx^n}$(sinx) = sin($\frac{n\pi}{2}$ +x).
- $\dfrac{d^n}{dx^n}$(cosx) = cos($\frac{n\pi}{2}$ +x).

Table of Contents

## nth Derivative of sinx

**Question:** Find the n-th derivative of sinx.

**Answer:**

Let y = sinx.

Then its first order derivative y_{1} is given as follows:

y_{1} = cos x

⇒ y_{1} = sin($\frac{\pi}{2}$ +x), by the rule sin($\frac{\pi}{2}$ +θ) = cosθ.

Differentiating y_{1} with respect to x, we get that

y_{2} = cos($\frac{\pi}{2}$ +x)⇒ y _{2} = sin($\frac{\pi}{2} +\frac{\pi}{2}$ +x) since sin($\frac{\pi}{2}$ +θ) = cosθ⇒ y _{2} = sin($2 \cdot \frac{\pi}{2}$ +x) |

Again differentiating y_{2} with respect to x, it follows that

y_{3} = cos($2 \cdot \frac{\pi}{2}$ +x)⇒ y _{3} = sin($\frac{\pi}{2} + 2 \cdot \frac{\pi}{2}$ +x), again by the same rule: sin($\frac{\pi}{2}$ +θ) = cosθ⇒ y _{3} = sin($3 \cdot \frac{\pi}{2}$ +x). |

Continuing this way, we observe that the n-th derivative of sinx will be given by

y_{n} = sin($n \cdot \frac{\pi}{2}$ +x).

**Remark:** The nth derivative of sin(ax) is equal to a^{n} sin($n \cdot \frac{\pi}{2}$ +ax).

**Also Read:** nth derivative of 1/x and 1/(ax+b)

Leibnitz Theorem on Successive Differentiation: Solved Problems

## nth Derivative of cosx

**Question:** Find the n-th derivative of cosx.

**Answer:**

Let y = cosx.

Then its first order derivative y_{1} is given as follows:

y_{1} = -sinx

⇒ y_{1} = cos($\frac{\pi}{2}$ +x), using the rule cos($\frac{\pi}{2}$ +θ) = -sinθ.

Differentiating y_{1} with respect to x, we get that

y_{2} = -sin($\frac{\pi}{2}$ +x)⇒ y _{2} = cos($\frac{\pi}{2} +\frac{\pi}{2}$ +x), this is because cos($\frac{\pi}{2}$ +θ) = -sinθ⇒ y _{2} = cos($2 \cdot \frac{\pi}{2}$ +x) |

Again differentiating y_{2} with respect to x, it follows that

y_{3} = -sin($2 \cdot \frac{\pi}{2}$ +x)⇒ y _{3} = cos($\frac{\pi}{2} + 2 \cdot \frac{\pi}{2}$ +x), again by the same rule: cos($\frac{\pi}{2}$ +θ) = -sinθ⇒ y _{3} = cos($3 \cdot \frac{\pi}{2}$ +x). |

Observing the pattern, we see that the n-th derivative of cosx will be given by

y_{n} = cos($n \cdot \frac{\pi}{2}$ +x).

**Remark:** The nth derivative of cos(ax) is equal to a^{n} cos($n \cdot \frac{\pi}{2}$ +ax).

## FAQs

**Q1: What is the n-th derivative of sin x?**

Answer: The n-th order derivative of sin x is equal to sin(nπ/2 +x).

**Q1: What is the n-th derivative of cos x?**

Answer: The n-th order derivative of cos x is equal to cos(nπ/2 +x).

This article is written by Dr. T. Mandal, Ph.D in Mathematics. On Mathstoon.com you will find Maths from very basic level to advanced level. Thanks for visiting.