The derivative of the sinc function is given by
The Correct Answer and Explanation is:
The derivative of the sinc function, defined as:sinc(x)=sinxx,x≠0,and sinc(0)=1\text{sinc}(x) = \frac{\sin x}{x}, \quad x \neq 0, \quad \text{and } \text{sinc}(0) = 1sinc(x)=xsinx,x=0,and sinc(0)=1
is given by:ddx[sinc(x)]=xcosx−sinxx2\frac{d}{dx}[\text{sinc}(x)] = \frac{x \cos x – \sin x}{x^2}dxd[sinc(x)]=x2xcosx−sinx
Explanation:
The sinc function arises frequently in signal processing and mathematical analysis, especially in Fourier transform theory. The function is defined piecewise to ensure continuity at x=0x = 0x=0. Although the expression sinxx\frac{\sin x}{x}xsinx is undefined at zero, the limit exists and is equal to 1. Therefore, sinc(0)=1\text{sinc}(0) = 1sinc(0)=1 is defined by continuity.
To differentiate sinc(x)=sinxx\text{sinc}(x) = \frac{\sin x}{x}sinc(x)=xsinx for x≠0x \neq 0x=0, the quotient rule is used. The quotient rule states that:ddx(u(x)v(x))=u′(x)v(x)−u(x)v′(x)[v(x)]2\frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) – u(x)v'(x)}{[v(x)]^2}dxd(v(x)u(x))=[v(x)]2u′(x)v(x)−u(x)v′(x)
Letting:
- u(x)=sinxu(x) = \sin xu(x)=sinx
- v(x)=xv(x) = xv(x)=x
Then:
- u′(x)=cosxu'(x) = \cos xu′(x)=cosx
- v′(x)=1v'(x) = 1v′(x)=1
Applying the quotient rule:ddx(sinxx)=cosx⋅x−sinx⋅1×2=xcosx−sinxx2\frac{d}{dx} \left( \frac{\sin x}{x} \right) = \frac{\cos x \cdot x – \sin x \cdot 1}{x^2} = \frac{x \cos x – \sin x}{x^2}dxd(xsinx)=x2cosx⋅x−sinx⋅1=x2xcosx−sinx
This derivative is undefined at x=0x = 0x=0, but the limit exists and can be evaluated using L’Hôpital’s Rule or series expansion to find the derivative at zero if needed.
The derivative of the sinc function reflects the combined oscillatory and decaying behavior of sinxx\frac{\sin x}{x}xsinx. Near the origin, sinc is smooth and well-behaved, and its derivative approaches 0, consistent with the maximum at x=0x = 0x=0.
