{"id":270735,"date":"2025-07-25T09:29:39","date_gmt":"2025-07-25T09:29:39","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=270735"},"modified":"2025-07-25T09:29:42","modified_gmt":"2025-07-25T09:29:42","slug":"calculate-the-derivative-of-the-function-ylog2-x-ln-2x-at-x1","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/25\/calculate-the-derivative-of-the-function-ylog2-x-ln-2x-at-x1\/","title":{"rendered":"Calculate the derivative of the function y=log2 x ln (2x) at x=1."},"content":{"rendered":"\n<p>Calculate the derivative of the function y=log2 x ln (2x) at x=1.<\/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>To find the derivative of the function y=log\u20612x\u22c5ln\u2061(2x)y = \\log_2 x \\cdot \\ln(2x)y=log2\u200bx\u22c5ln(2x) at x=1x = 1x=1, we need to apply both the product rule and the chain rule.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Apply the product rule<\/h3>\n\n\n\n<p>The product rule states that if we have two functions u(x)u(x)u(x) and v(x)v(x)v(x), then their derivative is given by:ddx[u(x)\u22c5v(x)]=u\u2032(x)\u22c5v(x)+u(x)\u22c5v\u2032(x)\\frac{d}{dx} [u(x) \\cdot v(x)] = u'(x) \\cdot v(x) + u(x) \\cdot v'(x)dxd\u200b[u(x)\u22c5v(x)]=u\u2032(x)\u22c5v(x)+u(x)\u22c5v\u2032(x)<\/p>\n\n\n\n<p>In our case, let:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>u(x)=log\u20612xu(x) = \\log_2 xu(x)=log2\u200bx<\/li>\n\n\n\n<li>v(x)=ln\u2061(2x)v(x) = \\ln(2x)v(x)=ln(2x)<\/li>\n<\/ul>\n\n\n\n<p>We will differentiate each of these separately.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Differentiate u(x)=log\u20612xu(x) = \\log_2 xu(x)=log2\u200bx<\/h3>\n\n\n\n<p>To differentiate u(x)=log\u20612xu(x) = \\log_2 xu(x)=log2\u200bx, we can use the change of base formula, which states:log\u20612x=ln\u2061xln\u20612\\log_2 x = \\frac{\\ln x}{\\ln 2}log2\u200bx=ln2lnx\u200b<\/p>\n\n\n\n<p>Now, differentiate using the chain rule:ddxlog\u20612x=1ln\u20612\u22c51x\\frac{d}{dx} \\log_2 x = \\frac{1}{\\ln 2} \\cdot \\frac{1}{x}dxd\u200blog2\u200bx=ln21\u200b\u22c5x1\u200b<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Differentiate v(x)=ln\u2061(2x)v(x) = \\ln(2x)v(x)=ln(2x)<\/h3>\n\n\n\n<p>For v(x)=ln\u2061(2x)v(x) = \\ln(2x)v(x)=ln(2x), we apply the chain rule. First, differentiate the outer function, which is ln\u2061(u)\\ln(u)ln(u), where u=2xu = 2xu=2x, and then multiply by the derivative of the inner function u=2xu = 2xu=2x:ddxln\u2061(2x)=12x\u22c52=1x\\frac{d}{dx} \\ln(2x) = \\frac{1}{2x} \\cdot 2 = \\frac{1}{x}dxd\u200bln(2x)=2&#215;1\u200b\u22c52=x1\u200b<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 4: Apply the product rule<\/h3>\n\n\n\n<p>Now, using the product rule:y\u2032(x)=1ln\u20612\u22c51x\u22c5ln\u2061(2x)+log\u20612x\u22c51xy'(x) = \\frac{1}{\\ln 2} \\cdot \\frac{1}{x} \\cdot \\ln(2x) + \\log_2 x \\cdot \\frac{1}{x}y\u2032(x)=ln21\u200b\u22c5x1\u200b\u22c5ln(2x)+log2\u200bx\u22c5x1\u200b<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 5: Evaluate at x=1x = 1x=1<\/h3>\n\n\n\n<p>Substitute x=1x = 1x=1 into the derivative:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>log\u206121=0\\log_2 1 = 0log2\u200b1=0 because log\u206121=0\\log_2 1 = 0log2\u200b1=0<\/li>\n\n\n\n<li>ln\u2061(2\u22c51)=ln\u2061(2)\\ln(2 \\cdot 1) = \\ln(2)ln(2\u22c51)=ln(2)<\/li>\n\n\n\n<li>11=1\\frac{1}{1} = 111\u200b=1<\/li>\n<\/ul>\n\n\n\n<p>Thus, the derivative at x=1x = 1x=1 becomes:y\u2032(1)=1ln\u20612\u22c5ln\u20612+0\u22c51=1y'(1) = \\frac{1}{\\ln 2} \\cdot \\ln 2 + 0 \\cdot 1 = 1y\u2032(1)=ln21\u200b\u22c5ln2+0\u22c51=1<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answer:<\/h3>\n\n\n\n<p>The derivative of y=log\u20612xln\u2061(2x)y = \\log_2 x \\ln(2x)y=log2\u200bxln(2x) at x=1x = 1x=1 is 111.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner6-1979.jpeg\" alt=\"\" class=\"wp-image-270740\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Calculate the derivative of the function y=log2 x ln (2x) at x=1. The Correct Answer and Explanation is: To find the derivative of the function y=log\u20612x\u22c5ln\u2061(2x)y = \\log_2 x \\cdot \\ln(2x)y=log2\u200bx\u22c5ln(2x) at x=1x = 1x=1, we need to apply both the product rule and the chain rule. Step 1: Apply the product rule The product [&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 center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-270735","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/270735","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=270735"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/270735\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=270735"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=270735"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=270735"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}