nth Derivative of sinx | nth Derivative of cosx

The nth 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).

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

y_n = 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

y_n = 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).

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