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 nth 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 y1 is given as follows:
y1 = cos x
⇒ y1 = sin($\frac{\pi}{2}$ +x), by the rule sin($\frac{\pi}{2}$ +θ) = cosθ.
Differentiating y1 with respect to x, we get that
y2 = cos($\frac{\pi}{2}$ +x) ⇒ y2 = sin($\frac{\pi}{2} +\frac{\pi}{2}$ +x) since sin($\frac{\pi}{2}$ +θ) = cosθ ⇒ y2 = sin($2 \cdot \frac{\pi}{2}$ +x) |
Again differentiating y2 with respect to x, it follows that
y3 = cos($2 \cdot \frac{\pi}{2}$ +x) ⇒ y3 = sin($\frac{\pi}{2} + 2 \cdot \frac{\pi}{2}$ +x), again by the same rule: sin($\frac{\pi}{2}$ +θ) = cosθ ⇒ y3 = sin($3 \cdot \frac{\pi}{2}$ +x). |
Continuing this way, we observe that the n-th derivative of sinx will be given by
yn = sin($n \cdot \frac{\pi}{2}$ +x).
Remark: The nth derivative of sin(ax) is equal to an 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 y1 is given as follows:
y1 = -sinx
⇒ y1 = cos($\frac{\pi}{2}$ +x), using the rule cos($\frac{\pi}{2}$ +θ) = -sinθ.
Differentiating y1 with respect to x, we get that
y2 = -sin($\frac{\pi}{2}$ +x) ⇒ y2 = cos($\frac{\pi}{2} +\frac{\pi}{2}$ +x), this is because cos($\frac{\pi}{2}$ +θ) = -sinθ ⇒ y2 = cos($2 \cdot \frac{\pi}{2}$ +x) |
Again differentiating y2 with respect to x, it follows that
y3 = -sin($2 \cdot \frac{\pi}{2}$ +x) ⇒ y3 = cos($\frac{\pi}{2} + 2 \cdot \frac{\pi}{2}$ +x), again by the same rule: cos($\frac{\pi}{2}$ +θ) = -sinθ ⇒ y3 = cos($3 \cdot \frac{\pi}{2}$ +x). |
Observing the pattern, we see that the n-th derivative of cosx will be given by
yn = cos($n \cdot \frac{\pi}{2}$ +x).
Remark: The nth derivative of cos(ax) is equal to an cos($n \cdot \frac{\pi}{2}$ +ax).
FAQs
Answer: The n-th order derivative of sin x is equal to sin(nπ/2 +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.