{"@context": "https://schema.org/", "@type": "Quiz", "about": {"@type": "Thing", "name": "Geometrical Progression" }, "educationalAlignment": [{"@type": "AlignmentObject","alignmentType": "educationalSubject","targetName": "Mathematics"}], "hasPart": [{"@context": "https://schema.org/","@type": "Question","eduQuestionType": "Flashcard", "text": "Find y'(0) if y=(x+1)(x+2)(x+3)...(x+2023) - ReadAxis", "acceptedAnswer": {"@type": "Answer", "text": "lny=ln(x+1)+ln(x+2)+ln(x+3)+…..+ln(x+2023)Differentiating with respect to xd(lny)dx=ln(x+1)+ln(x+2)+ln(x+3)+…..+ln(x+2023)dx⇒1y⋅dydx=1x+1⋅(0+1)+1x+2⋅(0+1)+1x+3⋅(0+1)+….+1x+2023⋅(0+1)⇒dydx=y′(x)=y(1x+1+1x+2+1x+3+…..+1x+2023)⇒y′(x)=(x+1)(x+2)(x+3)…..(x+2023)(1x+1+1x+2+1x+3+…..+1x+2023)Therefore y′(0)=1⋅2⋅3…..2023(11+12+13+…..+12023)⇒y′(0)=2023!⋅(11+12+13+…..+12023)" }}]} {"@context": "http://schema.org","@type": "QAPage", "name": "Find y'(0) if y=(x+1)(x+2)(x+3)...(x+2023) - ReadAxis", "description": "lny=ln(x+1)+ln(x+2)+ln(x+3)+…..+ln(x+2023)Differentiating with respect to xd(lny)dx=ln(x+1)+ln(x+2)+ln(x+3)+…..+ln(x+2023)dx⇒1y⋅dydx=1x+1⋅(0+1)+1x+2⋅(0+1)+1x+3⋅(0+1)+….+1x+2023⋅(0+1)⇒dydx=y′(x)=y(1x+1+1x+2+1x+3+…..+1x+2023)⇒y′(x)=(x+1)(x+2)(x+3)…..(x+2023)(1x+1+1x+2+1x+3+…..+1x+2023)Therefore y′(0)=1⋅2⋅3…..2023(11+12+13+…..+12023)⇒y′(0)=2023!⋅(11+12+13+…..+12023)", "mainEntity": {"@type": "Question", "@id": "https://readaxis.com/", "name": "Find y'(0) if y=(x+1)(x+2)(x+3)...(x+2023) - ReadAxis", "text": "Find y'(0) if y=(x+1)(x+2)(x+3)...(x+2023) - ReadAxis", "dateCreated": "2023-01-20T09:21:12.381Z", "answerCount": "1", "author": { "@type": "Person", "name": "ReadAxis", "url": "https://www.readaxis.com/" }, "acceptedAnswer": { "@type": "Answer", "upvoteCount": "4", "text": "lny=ln(x+1)+ln(x+2)+ln(x+3)+…..+ln(x+2023)Differentiating with respect to xd(lny)dx=ln(x+1)+ln(x+2)+ln(x+3)+…..+ln(x+2023)dx⇒1y⋅dydx=1x+1⋅(0+1)+1x+2⋅(0+1)+1x+3⋅(0+1)+….+1x+2023⋅(0+1)⇒dydx=y′(x)=y(1x+1+1x+2+1x+3+…..+1x+2023)⇒y′(x)=(x+1)(x+2)(x+3)…..(x+2023)(1x+1+1x+2+1x+3+…..+1x+2023)Therefore y′(0)=1⋅2⋅3…..2023(11+12+13+…..+12023)⇒y′(0)=2023!⋅(11+12+13+…..+12023)", "url": "https://readaxis.com/", "dateCreated": "2023-01-20T09:21:12.381Z", "author": { "@type": "Person", "name": "Readaxis Admin" } }, "suggestedAnswer": [] } }

Question

Find \(y'(0)\) if \(y = (x+1) (x+2) (x+3) … (x+2023)\)

Solution:

\(\ln y = \ln (x+1) + \ln (x+2) + \ln (x+3) + ….. + \ln (x+2023)\)

Differentiating with respect to \(x\)

\(\frac{d(\ln y)}{dx} = \frac{\ln (x+1) + \ln (x+2) + \ln (x+3) + ….. + \ln (x+2023)}{dx}\)

\(\Rightarrow \frac{1}{y}\cdot\frac{dy}{dx} = \frac{1}{x+1}\cdot(0 + 1) + \frac{1}{x+2}\cdot(0 + 1) + \frac{1}{x+3}\cdot(0 + 1) + …. + \frac{1}{x+2023}\cdot(0 + 1)\)

\(\Rightarrow \frac{dy}{dx} = y'(x) = y(\frac{1}{x+1} + \frac{1}{x+2} + \frac{1}{x+3} + ….. + \frac{1}{x+2023})\)

\(\Rightarrow y'(x) = (x+1)(x+2)(x+3)…..(x+2023)(\frac{1}{x+1} + \frac{1}{x+2} + \frac{1}{x+3} + ….. + \frac{1}{x+2023})\)

Therefore \(y'(0) = 1\cdot 2 \cdot 3 ….. 2023 (\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + ….. + \frac{1}{2023})\)

\(\Rightarrow y'(0) = 2023! \cdot (\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + ….. + \frac{1}{2023})\)