What is the nth Derivative of sinx? [Solved]

The nth derivative of sinx is equal to sin(nπ/2 +x). The nth derivative of sin x is denoted by dⁿ/dxⁿ (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

y₁ = cosx.

⇒ y₁ = sin$\left(\dfrac{\pi}{2}+x \right)$ using the trigonometric formula sin(π/2 +θ) = cosθ.

Differentiating y₁ with respect to x, we obtain that

y₂ = cos$\left(\dfrac{\pi}{2}+x \right)$.

⇒ y₂ = sin$\left(\dfrac{\pi}{2}+\dfrac{\pi}{2}+x \right)$ by the above rule: sin(π/2 +θ) = cosθ.

⇒ y₂ = sin$\left(\dfrac{2\pi}{2}+x \right)$.

In a similar way as above, it follows that

y₃ = cos$\left(\dfrac{2\pi}{2}+x \right)$ = sin$\left(\dfrac{3\pi}{2}+x \right)$.

y₄ = 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, dⁿ/dxⁿ (sinx) = sin(nπ/2 +x).

Also Read:

nth derivative of 1/x	nth derivative of xn
nth derivative of 1/(ax+b)

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 2ⁿ sin(nπ/2 +2x).

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