Find nth Derivative of sinx and cosx

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).

nth Derivative of sinx

Question: Find the n-th derivative of sinx.

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.

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

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).

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