{"id":47897,"date":"2024-09-04T15:55:23","date_gmt":"2024-09-04T15:55:23","guid":{"rendered":"https:\/\/www.oracletutoring.ca\/blog\/?p=47897"},"modified":"2024-09-04T15:55:23","modified_gmt":"2024-09-04T15:55:23","slug":"calculus-trig-derivative-product-rule-or-identity-substitution","status":"publish","type":"post","link":"https:\/\/www.oracletutoring.ca\/blog\/calculus-trig-derivative-product-rule-or-identity-substitution\/","title":{"rendered":"Calculus: trig derivative: product rule or identity substitution?"},"content":{"rendered":"\n<h2>Tutoring calculus, trig identities can give choices when performing derivatives. The tutor mentions an example.<\/h2>\n<p style=\"font-weight:bold\">\nExample: find derivative of y=sinxcosx\n<\/p>\n<p>\nSolution method 1: product rule (see my post <a href=\"https:\/\/www.oracletutoring.ca\/blog\/calculus-how-to-use-the-product-rule\/\">here.)<\/a><\/p>\n<p>\ny&#8217; = (sinx)&#8217;cosx + sinx(cosx)&#8217; = cosxcosx &#8211; sinxsinx = (cosx)^2 &#8211; (sinx)^2<\/p>\n<p>\nSolution method 2: identity substitution (sinxcosx= (sin2x)\/2) with chain rule\n<\/p>\n<p>\ny&#8217; = [(sin2x)\/2]&#8217; = (1\/2)(sin2x)&#8217; = (1\/2)(cos2x)(2) = cos2x<\/p>\n<p>\nNote that (cosx)^2 &#8211; (sinx)^2 = cos2x<\/p>\n<p>\nSource:<\/p>\n<p><a href=\"https:\/\/www2.clarku.edu\/faculty\/djoyce\/trig\/identities.html\">clarku.edu<\/a><\/p>\n<p>Larson, Roland E. and Robert P. Hostetler. <em>Calculus, Part One, Third Edition<\/em>. Toronto: D. C. Heath and Company, 1989.<\/p>\nJack of <a href=\"https:\/\/www.oracletutoring.ca\">Oracle Tutoring by Jack and Diane,<\/a> Campbell River, BC.\n","protected":false},"excerpt":{"rendered":"<p>Tutoring calculus, trig identities can give choices when performing derivatives. The tutor mentions an example. Example: find derivative of y=sinxcosx Solution method 1: product rule (see my post here.) y&#8217; = (sinx)&#8217;cosx + sinx(cosx)&#8217; = cosxcosx &#8211; sinxsinx = (cosx)^2 &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"more-link\" href=\"https:\/\/www.oracletutoring.ca\/blog\/calculus-trig-derivative-product-rule-or-identity-substitution\/\"> <span class=\"screen-reader-text\">Calculus: trig derivative: product rule or identity substitution?<\/span> Read More &raquo;<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[234],"tags":[301,76,3351,3350],"class_list":["post-47897","post","type-post","status-publish","format-standard","hentry","category-calculus","tag-product-rule","tag-substitution","tag-trig-derivative","tag-trig-identity"],"_links":{"self":[{"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/posts\/47897","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/comments?post=47897"}],"version-history":[{"count":5,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/posts\/47897\/revisions"}],"predecessor-version":[{"id":47902,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/posts\/47897\/revisions\/47902"}],"wp:attachment":[{"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/media?parent=47897"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/categories?post=47897"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/tags?post=47897"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}