{"id":33608,"date":"2025-06-23T09:47:38","date_gmt":"2025-06-23T09:47:38","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=33608"},"modified":"2025-06-23T09:47:39","modified_gmt":"2025-06-23T09:47:39","slug":"1-point-suppose-that-fx-te-exe-find-f-3-f3","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/1-point-suppose-that-fx-te-exe-find-f-3-f3\/","title":{"rendered":"(1 point) Suppose that f(x) = Te&#8217; exe Find f &#8216; (3). f'(3)"},"content":{"rendered":"\n<p>(1 point) Suppose that f(x) = Te&#8217; exe Find f &#8216; (3). f'(3)<\/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 solve for f\u2032(3)f'(3)f\u2032(3), let&#8217;s go through the problem step by step.<\/p>\n\n\n\n<p>We are given:<br><strong>f(x)=7ex+xexf(x) = 7e^x + xe^xf(x)=7ex+xex<\/strong><\/p>\n\n\n\n<p>We need to find the derivative f\u2032(x)f'(x)f\u2032(x), then evaluate it at x=3x = 3x=3.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Find the derivative f\u2032(x)f'(x)f\u2032(x)<\/h3>\n\n\n\n<p>The function f(x)f(x)f(x) has two terms: 7ex7e^x7ex and xexxe^xxex.<br>We will apply basic differentiation rules:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The derivative of exe^xex is exe^xex<\/li>\n\n\n\n<li>The derivative of a constant multiplied by a function follows the constant rule<\/li>\n\n\n\n<li>For xexxe^xxex, we use the product rule:<br>ddx(uv)=u\u2032v+uv\u2032\\frac{d}{dx} (uv) = u&#8217;v + uv&#8217;dxd\u200b(uv)=u\u2032v+uv\u2032<\/li>\n<\/ul>\n\n\n\n<p><strong>Differentiate each term:<\/strong><\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>ddx(7ex)=7ex\\frac{d}{dx} (7e^x) = 7e^xdxd\u200b(7ex)=7ex<\/li>\n\n\n\n<li>ddx(xex)\\frac{d}{dx} (xe^x)dxd\u200b(xex) using product rule:\n<ul class=\"wp-block-list\">\n<li>u=xu = xu=x, so u\u2032=1u&#8217; = 1u\u2032=1<\/li>\n\n\n\n<li>v=exv = e^xv=ex, so v\u2032=exv&#8217; = e^xv\u2032=ex<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n<p>Thus:<br>ddx(xex)=(1)(ex)+(x)(ex)=ex+xex\\frac{d}{dx} (xe^x) = (1)(e^x) + (x)(e^x) = e^x + xe^xdxd\u200b(xex)=(1)(ex)+(x)(ex)=ex+xex<\/p>\n\n\n\n<p><strong>So, the full derivative is:<\/strong><br>f\u2032(x)=7ex+ex+xexf'(x) = 7e^x + e^x + xe^xf\u2032(x)=7ex+ex+xex<\/p>\n\n\n\n<p>We can factor terms:<br>f\u2032(x)=7ex+ex+xex=(7ex+ex)+xex=8ex+xexf'(x) = 7e^x + e^x + xe^x = (7e^x + e^x) + xe^x = 8e^x + xe^xf\u2032(x)=7ex+ex+xex=(7ex+ex)+xex=8ex+xex<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Evaluate f\u2032(3)f'(3)f\u2032(3)<\/h3>\n\n\n\n<p>We substitute x=3x = 3x=3 into the derivative:<\/p>\n\n\n\n<p>f\u2032(3)=8e3+3e3=(8+3)e3=11e3f'(3) = 8e^3 + 3e^3 = (8 + 3)e^3 = 11e^3f\u2032(3)=8e3+3e3=(8+3)e3=11e3<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answer:<\/h3>\n\n\n\n<p><strong>f\u2032(3)=11e3f'(3) = 11e^3f\u2032(3)=11e3<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<p>In this problem, we apply fundamental differentiation rules: constant multiples, the derivative of the exponential function, and the product rule. The first term 7ex7e^x7ex differentiates directly to 7ex7e^x7ex, while the second term xexxe^xxex requires the product rule, yielding ex+xexe^x + xe^xex+xex. Combining terms gives 8ex+xex8e^x + xe^x8ex+xex. Plugging in x=3x = 3x=3 provides the final result of 11e311e^311e3, representing the slope of the tangent line to the function f(x)f(x)f(x) at the point where x=3x = 3x=3. This derivative indicates how rapidly the function changes at that specific point.<\/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-banner8-530.jpeg\" alt=\"\" class=\"wp-image-33609\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-530.jpeg 852w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-530-250x300.jpeg 250w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-530-768x923.jpeg 768w\" sizes=\"auto, (max-width: 852px) 100vw, 852px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>(1 point) Suppose that f(x) = Te&#8217; exe Find f &#8216; (3). f'(3) The Correct Answer and Explanation is: To solve for f\u2032(3)f'(3)f\u2032(3), let&#8217;s go through the problem step by step. We are given:f(x)=7ex+xexf(x) = 7e^x + xe^xf(x)=7ex+xex We need to find the derivative f\u2032(x)f'(x)f\u2032(x), then evaluate it at x=3x = 3x=3. Step 1: Find [&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-33608","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/33608","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=33608"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/33608\/revisions"}],"predecessor-version":[{"id":33610,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/33608\/revisions\/33610"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=33608"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=33608"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=33608"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}