{"id":38263,"date":"2025-06-26T08:53:53","date_gmt":"2025-06-26T08:53:53","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=38263"},"modified":"2025-06-26T08:53:54","modified_gmt":"2025-06-26T08:53:54","slug":"find-the-derivative-ft-3t-sin-t","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/find-the-derivative-ft-3t-sin-t\/","title":{"rendered":"Find the derivative. f(t) = 3t sin? t"},"content":{"rendered":"\n<p>Find the derivative. f(t) = 3t sin? t<\/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>To find the derivative of the function:<br><strong>f(t) = 3t \u00b7 sin\u00b2(t)<\/strong><\/p>\n\n\n\n<p>First, rewrite the function using exponent notation:<br><strong>f(t) = 3t \u00b7 (sin(t))\u00b2<\/strong><\/p>\n\n\n\n<p>Now apply the <strong>product rule<\/strong>, since the function is a product of two functions:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>First function: <strong>u = 3t<\/strong><\/li>\n\n\n\n<li>Second function: <strong>v = (sin(t))\u00b2<\/strong><\/li>\n<\/ul>\n\n\n\n<p>The <strong>product rule<\/strong> is:<br><strong>f'(t) = u&#8217;v + uv&#8217;<\/strong><\/p>\n\n\n\n<p>Let us differentiate each part.<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>u = 3t \u2192 u&#8217; = 3<\/strong><\/li>\n\n\n\n<li><strong>v = (sin(t))\u00b2<\/strong><br>Use the <strong>chain rule<\/strong>:\n<ul class=\"wp-block-list\">\n<li>The outer function is something squared, so its derivative is 2\u00b7(sin(t))<\/li>\n\n\n\n<li>The inner function is sin(t), whose derivative is cos(t)<br>So:<br><strong>v&#8217; = 2\u00b7sin(t)\u00b7cos(t)<\/strong><\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n<p>Now apply the product rule:<br><strong>f'(t) = 3 \u00b7 (sin(t))\u00b2 + 3t \u00b7 2\u00b7sin(t)\u00b7cos(t)<\/strong><\/p>\n\n\n\n<p>Simplify the expression:<br><strong>f'(t) = 3(sin(t))\u00b2 + 6t\u00b7sin(t)\u00b7cos(t)<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answer:<\/h3>\n\n\n\n<p><strong>f'(t) = 3(sin(t))\u00b2 + 6t\u00b7sin(t)\u00b7cos(t)<\/strong><\/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>To find the derivative of a function like <strong>f(t) = 3t \u00b7 sin\u00b2(t)<\/strong>, you must recognize that it is the product of two functions: one involving <strong>t<\/strong> and one involving <strong>sin(t)<\/strong>. This calls for using the <strong>product rule<\/strong>, which is a fundamental rule in calculus for differentiating the product of two functions.<\/p>\n\n\n\n<p>The product rule says that if <strong>f(t) = u(t) \u00b7 v(t)<\/strong>, then its derivative is <strong>f'(t) = u'(t) \u00b7 v(t) + u(t) \u00b7 v'(t)<\/strong>. In this problem, <strong>u(t)<\/strong> is the linear function <strong>3t<\/strong>, and <strong>v(t)<\/strong> is the trigonometric function squared, <strong>(sin(t))\u00b2<\/strong>.<\/p>\n\n\n\n<p>Differentiating <strong>u(t) = 3t<\/strong> is straightforward, giving <strong>u'(t) = 3<\/strong>.<\/p>\n\n\n\n<p>For the <strong>v(t) = (sin(t))\u00b2<\/strong>, you use the chain rule. The outer function is a square, and the inner function is <strong>sin(t)<\/strong>. The derivative of <strong>(sin(t))\u00b2<\/strong> becomes <strong>2\u00b7sin(t)\u00b7cos(t)<\/strong>.<\/p>\n\n\n\n<p>Putting it all together using the product rule:<br><strong>f'(t) = 3\u00b7(sin(t))\u00b2 + 3t\u00b72\u00b7sin(t)\u00b7cos(t)<\/strong><br>This simplifies to:<br><strong>f'(t) = 3(sin(t))\u00b2 + 6t\u00b7sin(t)\u00b7cos(t)<\/strong><\/p>\n\n\n\n<p>This expression gives the rate of change of the original function, combining both linear and trigonometric components.<\/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-banner8-1014.jpeg\" alt=\"\" class=\"wp-image-38267\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1014.jpeg 852w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1014-250x300.jpeg 250w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1014-768x923.jpeg 768w\" sizes=\"auto, (max-width: 852px) 100vw, 852px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Find the derivative. f(t) = 3t sin? t The Correct Answer and Explanation is: To find the derivative of the function:f(t) = 3t \u00b7 sin\u00b2(t) First, rewrite the function using exponent notation:f(t) = 3t \u00b7 (sin(t))\u00b2 Now apply the product rule, since the function is a product of two functions: The product rule is:f'(t) = [&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-38263","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/38263","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=38263"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/38263\/revisions"}],"predecessor-version":[{"id":38268,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/38263\/revisions\/38268"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=38263"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=38263"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=38263"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}