{"id":44707,"date":"2025-06-30T18:28:11","date_gmt":"2025-06-30T18:28:11","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=44707"},"modified":"2025-06-30T18:28:12","modified_gmt":"2025-06-30T18:28:12","slug":"i-need-help-finding-the-derivative-of-2sinx-because-i-keep-on-getting-2sinxcosx-instead-of-2cosx","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/i-need-help-finding-the-derivative-of-2sinx-because-i-keep-on-getting-2sinxcosx-instead-of-2cosx\/","title":{"rendered":"I need help finding the derivative of 2sin(x) because I keep on getting 2sin(x)cos(x) instead of 2cos(x)."},"content":{"rendered":"\n<p>I need help finding the derivative of 2sin(x) because I keep on getting 2sin(x)cos(x) instead of 2cos(x). Do I use the chain rule?<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>You&#8217;re on the right track, but it looks like you might be confusing the result for the derivative of sin\u2061(2x)\\sin(2x)sin(2x) with the derivative of 2sin\u2061(x)2\\sin(x)2sin(x). Let&#8217;s clarify the steps.<\/p>\n\n\n\n<p>To find the derivative of 2sin\u2061(x)2\\sin(x)2sin(x), you do <strong>not<\/strong> need the chain rule because there is no composite function like sin\u2061(2x)\\sin(2x)sin(2x) involved. Instead, you just apply basic differentiation rules.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Derivative of 2sin\u2061(x)2\\sin(x)2sin(x)<\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Constant Multiplier Rule<\/strong>: The constant multiple (in this case, 2) can be factored out of the derivative operation. So, you are left with: ddx[2sin\u2061(x)]=2\u22c5ddx[sin\u2061(x)]\\frac{d}{dx}[2\\sin(x)] = 2 \\cdot \\frac{d}{dx}[\\sin(x)]dxd\u200b[2sin(x)]=2\u22c5dxd\u200b[sin(x)]<\/li>\n\n\n\n<li><strong>Derivative of sin\u2061(x)\\sin(x)sin(x)<\/strong>: The derivative of sin\u2061(x)\\sin(x)sin(x) with respect to xxx is simply cos\u2061(x)\\cos(x)cos(x). Therefore: 2\u22c5cos\u2061(x)2 \\cdot \\cos(x)2\u22c5cos(x)<\/li>\n<\/ol>\n\n\n\n<p>Thus, the derivative of 2sin\u2061(x)2\\sin(x)2sin(x) is 2cos\u2061(x)2\\cos(x)2cos(x).<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Why not 2sin\u2061(x)cos\u2061(x)2\\sin(x)\\cos(x)2sin(x)cos(x)?<\/h3>\n\n\n\n<p>The expression 2sin\u2061(x)cos\u2061(x)2\\sin(x)\\cos(x)2sin(x)cos(x) is the result you would get if you were differentiating sin\u2061(2x)\\sin(2x)sin(2x) using the chain rule. Here\u2019s why:<\/p>\n\n\n\n<p>When you differentiate sin\u2061(2x)\\sin(2x)sin(2x), the chain rule tells you to first differentiate the outer function sin\u2061(u)\\sin(u)sin(u), where u=2xu = 2xu=2x, and then multiply by the derivative of 2x2x2x. This process leads to the following:ddx[sin\u2061(2x)]=cos\u2061(2x)\u22c5ddx[2x]=2cos\u2061(2x)\\frac{d}{dx}[\\sin(2x)] = \\cos(2x) \\cdot \\frac{d}{dx}[2x] = 2\\cos(2x)dxd\u200b[sin(2x)]=cos(2x)\u22c5dxd\u200b[2x]=2cos(2x)<\/p>\n\n\n\n<p>But that&#8217;s not the case here. Since you&#8217;re differentiating 2sin\u2061(x)2\\sin(x)2sin(x), you only need to apply the basic rule for the derivative of sin\u2061(x)\\sin(x)sin(x), resulting in 2cos\u2061(x)2\\cos(x)2cos(x).<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Conclusion<\/h3>\n\n\n\n<p>The correct derivative of 2sin\u2061(x)2\\sin(x)2sin(x) is 2cos\u2061(x)2\\cos(x)2cos(x), and no chain rule is needed. You only use the chain rule when differentiating composite functions.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"852\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1534.jpeg\" alt=\"\" class=\"wp-image-44708\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1534.jpeg 852w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1534-250x300.jpeg 250w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1534-768x923.jpeg 768w\" sizes=\"auto, (max-width: 852px) 100vw, 852px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>I need help finding the derivative of 2sin(x) because I keep on getting 2sin(x)cos(x) instead of 2cos(x). Do I use the chain rule? The Correct Answer and Explanation is: You&#8217;re on the right track, but it looks like you might be confusing the result for the derivative of sin\u2061(2x)\\sin(2x)sin(2x) with the derivative of 2sin\u2061(x)2\\sin(x)2sin(x). Let&#8217;s [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-44707","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/44707","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=44707"}],"version-history":[{"count":2,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/44707\/revisions"}],"predecessor-version":[{"id":44710,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/44707\/revisions\/44710"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=44707"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=44707"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=44707"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}