{"id":232190,"date":"2025-06-11T19:13:16","date_gmt":"2025-06-11T19:13:16","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=232190"},"modified":"2025-06-11T19:13:18","modified_gmt":"2025-06-11T19:13:18","slug":"the-derivative-of-the-sinc-function-is-given-by","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/11\/the-derivative-of-the-sinc-function-is-given-by\/","title":{"rendered":"The derivative of the sinc function is given by"},"content":{"rendered":"\n<p>The derivative of the sinc function is given by<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>The derivative of the <strong>sinc<\/strong> function, defined as:sinc(x)=sin\u2061xx,x\u22600,and&nbsp;sinc(0)=1\\text{sinc}(x) = \\frac{\\sin x}{x}, \\quad x \\neq 0, \\quad \\text{and } \\text{sinc}(0) = 1sinc(x)=xsinx\u200b,x\ue020=0,and&nbsp;sinc(0)=1<\/p>\n\n\n\n<p>is given by:ddx[sinc(x)]=xcos\u2061x\u2212sin\u2061xx2\\frac{d}{dx}[\\text{sinc}(x)] = \\frac{x \\cos x &#8211; \\sin x}{x^2}dxd\u200b[sinc(x)]=x2xcosx\u2212sinx\u200b<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<p>The <strong>sinc<\/strong> 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 sin\u2061xx\\frac{\\sin x}{x}xsinx\u200b 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.<\/p>\n\n\n\n<p>To differentiate sinc(x)=sin\u2061xx\\text{sinc}(x) = \\frac{\\sin x}{x}sinc(x)=xsinx\u200b for x\u22600x \\neq 0x\ue020=0, the <strong>quotient rule<\/strong> is used. The quotient rule states that:ddx(u(x)v(x))=u\u2032(x)v(x)\u2212u(x)v\u2032(x)[v(x)]2\\frac{d}{dx} \\left( \\frac{u(x)}{v(x)} \\right) = \\frac{u'(x)v(x) &#8211; u(x)v'(x)}{[v(x)]^2}dxd\u200b(v(x)u(x)\u200b)=[v(x)]2u\u2032(x)v(x)\u2212u(x)v\u2032(x)\u200b<\/p>\n\n\n\n<p>Letting:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>u(x)=sin\u2061xu(x) = \\sin xu(x)=sinx<\/li>\n\n\n\n<li>v(x)=xv(x) = xv(x)=x<\/li>\n<\/ul>\n\n\n\n<p>Then:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>u\u2032(x)=cos\u2061xu'(x) = \\cos xu\u2032(x)=cosx<\/li>\n\n\n\n<li>v\u2032(x)=1v'(x) = 1v\u2032(x)=1<\/li>\n<\/ul>\n\n\n\n<p>Applying the quotient rule:ddx(sin\u2061xx)=cos\u2061x\u22c5x\u2212sin\u2061x\u22c51&#215;2=xcos\u2061x\u2212sin\u2061xx2\\frac{d}{dx} \\left( \\frac{\\sin x}{x} \\right) = \\frac{\\cos x \\cdot x &#8211; \\sin x \\cdot 1}{x^2} = \\frac{x \\cos x &#8211; \\sin x}{x^2}dxd\u200b(xsinx\u200b)=x2cosx\u22c5x\u2212sinx\u22c51\u200b=x2xcosx\u2212sinx\u200b<\/p>\n\n\n\n<p>This derivative is undefined at x=0x = 0x=0, but the limit exists and can be evaluated using L&#8217;H\u00f4pital\u2019s Rule or series expansion to find the derivative at zero if needed.<\/p>\n\n\n\n<p>The derivative of the sinc function reflects the combined oscillatory and decaying behavior of sin\u2061xx\\frac{\\sin x}{x}xsinx\u200b. Near the origin, sinc is smooth and well-behaved, and its derivative approaches 0, consistent with the maximum at x=0x = 0x=0.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-306.jpeg\" alt=\"\" class=\"wp-image-232191\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>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)=sin\u2061xx,x\u22600,and&nbsp;sinc(0)=1\\text{sinc}(x) = \\frac{\\sin x}{x}, \\quad x \\neq 0, \\quad \\text{and } \\text{sinc}(0) = 1sinc(x)=xsinx\u200b,x\ue020=0,and&nbsp;sinc(0)=1 is given by:ddx[sinc(x)]=xcos\u2061x\u2212sin\u2061xx2\\frac{d}{dx}[\\text{sinc}(x)] = \\frac{x \\cos x &#8211; \\sin x}{x^2}dxd\u200b[sinc(x)]=x2xcosx\u2212sinx\u200b Explanation: The sinc function arises frequently in signal processing [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center 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