{"id":133954,"date":"2023-03-12T17:11:18","date_gmt":"2023-03-12T11:41:18","guid":{"rendered":"https:\/\/www.mapsofindia.com\/my-india\/?p=133954"},"modified":"2023-03-12T17:11:18","modified_gmt":"2023-03-12T11:41:18","slug":"if-ytan-xsec-xthen-prove-that-d2y-dx2-cos-x-1-sin-x2","status":"publish","type":"post","link":"https:\/\/www.mapsofindia.com\/my-india\/quiz\/if-ytan-xsec-xthen-prove-that-d2y-dx2-cos-x-1-sin-x2","title":{"rendered":"if y=tan x+sec x,then prove that d2y\/dx2 =cos x\/(1-sin x)2"},"content":{"rendered":"<h2>Question: if y=tan x+sec x,then prove that d2y\/dx2 =cos x\/(1-sin x)2<\/h2>\n<p>&nbsp;<\/p>\n<h3>The correct answer is &#8211;<\/h3>\n<div class=\"group w-full text-gray-800 dark:text-gray-100 border-b border-black\/10 dark:border-gray-900\/50 bg-gray-50 dark:bg-[#444654]\">\n<div class=\"text-base gap-4 md:gap-6 md:max-w-2xl lg:max-w-2xl xl:max-w-3xl p-4 md:py-6 flex lg:px-0 m-auto\">\n<div class=\"relative flex w-[calc(100%-50px)] flex-col gap-1 md:gap-3 lg:w-[calc(100%-115px)]\">\n<div class=\"flex flex-grow flex-col gap-3\">\n<div class=\"min-h-[20px] flex flex-col items-start gap-4 whitespace-pre-wrap\">\n<div class=\"markdown prose w-full break-words dark:prose-invert light\">\n<h4>To find the second derivative of y with respect to x, we first find the first derivative of y:<\/h4>\n<h4>dy\/dx = sec^2 x + tan x<\/h4>\n<h4>Next, we take the derivative of this expression with respect to x to find the second derivative:<\/h4>\n<h4>d2y\/dx2 = d\/dx (sec^2 x + tan x) = 2sec x sec x tan x + sec^2 x<\/h4>\n<h4>Now we use the identity sec^2 x &#8211; tan^2 x = 1 to write sec^2 x = tan^2 x + 1, and substitute this into our expression for the second derivative:<\/h4>\n<h4>d2y\/dx2 = 2sec x (tan^2 x + 1) tan x + (tan^2 x + 1) = 2tan x sec x + 2sec x tan^3 x + 1<\/h4>\n<h4>Next, we use the identity tan x = sin x \/ cos x and simplify the expression:<\/h4>\n<h4>d2y\/dx2 = 2(sin x \/ cos x)(1 \/ cos x) + 2(1 \/ cos x)(sin^3 x \/ cos x) + 1 = 2sin x \/ cos^2 x + 2sin^3 x \/ cos^3 x + 1 \/ cos^2 x<\/h4>\n<h4>We can then simplify this expression further by using the identity sin^2 x + cos^2 x = 1 to write sin^2 x = 1 &#8211; cos^2 x:<\/h4>\n<h4>d2y\/dx2 = 2sin x \/ cos^2 x + 2sin^3 x \/ cos^3 x + 1 \/ cos^2 x = 2sin x \/ cos^2 x + 2(sin^2 x)(sin x \/ cos^3 x) + 1 \/ cos^2 x = 2sin x \/ cos^2 x + 2(cos^2 x &#8211; 1)(sin x \/ cos^3 x) + 1 \/ cos^2 x = 2sin x \/ cos^2 x + 2sin x \/ cos x &#8211; 2 \/ cos^2 x + 1 \/ cos^2 x = 2sin x \/ cos^2 x + 2sin x \/ cos x &#8211; 1 \/ cos^2 x<\/h4>\n<h4>Finally, we use the identity 1 &#8211; sin^2 x = cos^2 x to write 1 \/ cos^2 x = 1 + tan^2 x, and substitute this into our expression:<\/h4>\n<h4>d2y\/dx2 = 2sin x \/ cos^2 x + 2sin x \/ cos x &#8211; 1 \/ cos^2 x = 2sin x \/ cos^2 x + 2sin x \/ cos x &#8211; 1 &#8211; tan^2 x = 2sin x \/ cos^2 x + 2sin x \/ cos x &#8211; sec^2 x<\/h4>\n<h4>We can then simplify this expression further by using the identity sin x = (1 \/ cos x)sin x, and factoring out (1 \/ cos x):<\/h4>\n<h4>d2y\/dx2 = 2sin x \/ cos^2 x + 2sin x \/ cos x &#8211; sec^2 x = 2sin x (1 + cos x) \/ cos^3 x &#8211; 1 \/ cos^2 x = (2sin x + 2sin x cos x &#8211; 1) \/ cos^3 x<\/h4>\n<h4>Next, we use the identity 1 &#8211; sin x cos x = sin^2 x \/ (1 + cos x) to write 2sin x cos x = 2sin<\/h4>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"flex justify-between\">\n<div class=\"text-gray-400 flex self-end lg:self-center justify-center mt-2 gap-3 md:gap-4 lg:gap-1 lg:absolute lg:top-0 lg:translate-x-full lg:right-0 lg:mt-0 lg:pl-2 visible\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Question: if y=tan x+sec x,then prove that d2y\/dx2 =cos x\/(1-sin x)2 &nbsp; The correct answer is &#8211; To find the second derivative of y with respect to x, we first find the first derivative of y: dy\/dx = sec^2 x + tan x Next, we take the derivative of this expression with respect to x [&hellip;]<\/p>\n","protected":false},"author":21842,"featured_media":132233,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[12119],"tags":[],"class_list":{"0":"post-133954","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-quiz"},"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/133954","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/users\/21842"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/comments?post=133954"}],"version-history":[{"count":1,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/133954\/revisions"}],"predecessor-version":[{"id":133955,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/133954\/revisions\/133955"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/media\/132233"}],"wp:attachment":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/media?parent=133954"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/categories?post=133954"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/tags?post=133954"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}