The nth derivative of sinx is equal to sin(nπ/2 +x). The nth derivative of sin x is denoted by d^{n}/dx^{n} (sinx), and its formula is given as follows:

$\boxed{\dfrac{d^n}{dx^n}\left( \sin x\right)=\sin \left(\dfrac{n \pi}{2}+x \right)}$

## nth Derivative of sin x

**Question:** Find the nth Derivative of sinx.

**Answer:**

To find the nth derivative of sinx with respect to x, let us put

y = sinx.

Its first derivative is given by

y_{1} = cosx.

⇒ y_{1} = sin$\left(\dfrac{\pi}{2}+x \right)$ using the trigonometric formula sin(π/2 +θ) = cosθ.

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

y_{2} = cos$\left(\dfrac{\pi}{2}+x \right)$.

⇒ y_{2} = sin$\left(\dfrac{\pi}{2}+\dfrac{\pi}{2}+x \right)$ by the above rule: sin(π/2 +θ) = cosθ.

⇒ y_{2} = sin$\left(\dfrac{2\pi}{2}+x \right)$.

In a similar way as above, it follows that

y_{3} = cos$\left(\dfrac{2\pi}{2}+x \right)$ = sin$\left(\dfrac{3\pi}{2}+x \right)$.

y_{4} = cos$\left(\dfrac{3\pi}{2}+x \right)$ = sin$\left(\dfrac{4\pi}{2}+x \right)$.

**Conclusion:** By observing the patterns, we see that the nth^{ }derivative of sinx is equal to sin(nπ/2 +x). That is, d^{n}/dx^{n} (sinx) = sin(nπ/2 +x).

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## Question-Answer

**Question 1:** Find the nth derivative of sin2x.

**Answer:**

As the nth^{ }derivative of sinx is equal to sin(nπ/2 +x), by the chain rule of differentiation the nth derivative of sin2x will be equal to 2^{n} sin(nπ/2 +2x).

## FAQs

### Q1: What is nth Derivative of sinx?

**Answer:** The nth derivative of sinx is equal to sin(nπ/2 +x).

### Q2: What is nth Derivative of sin2x?

**Answer:** The nth derivative of sin2x is equal to 2^{n} sin(nπ/2 +2x).

### Q3: What is nth Derivative of sin3x?

**Answer:** The nth derivative of sin3x is equal to 3^{n} sin(nπ/2 +3x).