Kisha saved $12.00 when buying a coat. The coat was on sale for 30% off. What was the original price of the coat?<\/p>\n\n\n\n
Differentiate the function.<\/p>\n\n\n\n
a. g(x) \u2013 In(x VP – 1).<\/p>\n\n\n\n
b. F(y) = y ln(1 + e’).<\/p>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n Let’s break your request into two parts:<\/p>\n\n\n\n Problem:<\/strong> Solution:<\/strong> 30% of x = $12 \u2705 Original price = $40.00<\/strong><\/p>\n\n\n\n Let\u2019s differentiate each function step-by-step and explain.<\/p>\n\n\n\n Note:<\/strong> It seems you meant a function like ln\u2061(xP\u22121)\\ln(x\\sqrt{P} – 1), assuming “VP” meant “\u221aP”. Let’s denote it clearly: g(x)=ln\u2061(xP\u22121)g(x) = \\ln(x\\sqrt{P} – 1)<\/p>\n\n\n\n Use the chain rule<\/strong>:<\/p>\n\n\n\n ddx[xP\u22121]=P\\frac{d}{dx}[x\\sqrt{P} – 1] = \\sqrt{P} g\u2032(x)=PxP\u22121g'(x) = \\frac{\\sqrt{P}}{x\\sqrt{P} – 1}<\/p>\n\n\n\n \u2705 Answer:<\/strong> g\u2032(x)=PxP\u22121g'(x) = \\frac{\\sqrt{P}}{x\\sqrt{P} – 1}<\/p>\n\n\n\n Use the product rule<\/strong>: F\u2032(y)=ddy[y]\u22c5ln\u2061(1+ey)+y\u22c5ddy[ln\u2061(1+ey)]F'(y) = \\frac{d}{dy}[y] \\cdot \\ln(1 + e^y) + y \\cdot \\frac{d}{dy}[\\ln(1 + e^y)]<\/p>\n\n\n\n Putting it all together: F\u2032(y)=ln\u2061(1+ey)+y\u22c5ey1+eyF'(y) = \\ln(1 + e^y) + y \\cdot \\frac{e^y}{1 + e^y}<\/p>\n\n\n\n \u2705 Answer:<\/strong> F\u2032(y)=ln\u2061(1+ey)+yey1+eyF'(y) = \\ln(1 + e^y) + \\frac{y e^y}{1 + e^y}<\/p>\n\n\n\n Differentiation is a core concept in calculus, used to determine how a function changes as its input changes. In both problems above, we use foundational rules of differentiation: the chain rule and product rule.<\/p>\n\n\n\n In function a<\/strong>, we\u2019re differentiating a natural logarithm function involving a linear term. The chain rule<\/strong> is essential when dealing with a composition of functions. For example, ln\u2061(u(x))\\ln(u(x)) requires applying the derivative of ln first, 1u\\frac{1}{u}, then multiplying it by the derivative of the inner function uu. This is why we took the derivative of xP\u22121x\\sqrt{P} – 1 (which is P\\sqrt{P}) and multiplied it by 1xP\u22121\\frac{1}{x\\sqrt{P} – 1}.<\/p>\n\n\n\n In function b<\/strong>, we\u2019re differentiating a product of two functions: yy and ln\u2061(1+ey)\\ln(1 + e^y). When differentiating products, we use the product rule<\/strong>, which says (uv)\u2032=u\u2032v+uv\u2032(uv)’ = u’v + uv’. The second part of the product includes a logarithmic function of an exponential expression, which itself requires the chain rule<\/strong>. The derivative of ln\u2061(1+ey)\\ln(1 + e^y) involves first differentiating the outer ln function, then multiplying by the derivative of the inner 1+ey1 + e^y, giving ey1+ey\\frac{e^y}{1 + e^y}.<\/p>\n\n\n\n Altogether, these problems test our ability to apply differentiation rules in a layered way. Recognizing which rules to apply \u2014 chain rule, product rule, and properties of logarithmic and exponential functions \u2014 is key to solving such calculus problems correctly.<\/p>\n","protected":false},"excerpt":{"rendered":" Kisha saved $12.00 when buying a coat. The coat was on sale for 30% off. What was the original price of the coat? Differentiate the function. a. g(x) \u2013 In(x VP – 1). b. F(y) = y ln(1 + e’). The Correct Answer and Explanation is: Let’s break your request into two parts: Part 1: […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-217112","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/217112","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=217112"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/217112\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=217112"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=217112"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=217112"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\nPart 1: Solving the Word Problem<\/strong><\/h3>\n\n\n\n
Kisha saved $12.00 with a 30% discount<\/strong> on a coat. What was the original price<\/strong>?<\/p>\n\n\n\n
Let the original price be x<\/strong>.<\/p>\n\n\n\n
\u2192 0.30x=120.30x = 12
\u2192 x=120.30=40x = \\frac{12}{0.30} = 40<\/p>\n\n\n\n
\n\n\n\nPart 2: Differentiating the Given Functions<\/strong><\/h3>\n\n\n\n
\n\n\n\na.<\/strong> g(x)=ln\u2061(x\u22c5P\u22121)g(x) = \\ln(x \\cdot \\sqrt{P} – 1)<\/h4>\n\n\n\n
\n
\n\n\n\nb.<\/strong> F(y)=yln\u2061(1+ey)F(y) = y \\ln(1 + e^y)<\/h4>\n\n\n\n
\n
\n\n\n\nExplanation (Approx. 300 Words)<\/strong><\/h3>\n\n\n\n