{"id":16780,"date":"2025-06-11T20:05:44","date_gmt":"2025-06-11T20:05:44","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=16780"},"modified":"2025-06-11T20:05:46","modified_gmt":"2025-06-11T20:05:46","slug":"differentiate-the-function-y-equals-left-parenthesis-9-x-superscript-4-baseline-minus-x-plus-8-right-parenthesis-left-parenthesis-negative-x-superscript-5-baseline-plus-2-right-parenthesis","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/differentiate-the-function-y-equals-left-parenthesis-9-x-superscript-4-baseline-minus-x-plus-8-right-parenthesis-left-parenthesis-negative-x-superscript-5-baseline-plus-2-right-parenthesis\/","title":{"rendered":"Differentiate the function y equals left parenthesis 9 x Superscript 4 Baseline minus x plus 8 right parenthesis left parenthesis negative x Superscript 5 Baseline plus 2 right parenthesis"},"content":{"rendered":"\n

Differentiate the function y equals left parenthesis 9 x Superscript 4 Baseline minus x plus 8 right parenthesis left parenthesis negative x Superscript 5 Baseline plus 2 right parenthesis<\/p>\n\n\n\n

The correct answer and explanation is:<\/strong><\/mark><\/p>\n\n\n\n

We are asked to differentiate the function: y=(9×4\u2212x+8)(\u2212x5+2)y = (9x^4 – x + 8)(-x^5 + 2)<\/p>\n\n\n\n

This is a product of two functions<\/strong>, so we will apply the Product Rule<\/strong> for differentiation.<\/p>\n\n\n\n

\n\n\n\n

Step 1: Recall the Product Rule<\/h3>\n\n\n\n

If: y=u(x)\u22c5v(x),y = u(x) \\cdot v(x),<\/p>\n\n\n\n

then: dydx=u\u2032(x)v(x)+u(x)v\u2032(x)\\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x)<\/p>\n\n\n\n

\n\n\n\n

Step 2: Define the two parts<\/h3>\n\n\n\n

Let:<\/p>\n\n\n\n

    \n
  • u(x)=9×4\u2212x+8u(x) = 9x^4 – x + 8<\/li>\n\n\n\n
  • v(x)=\u2212x5+2v(x) = -x^5 + 2<\/li>\n<\/ul>\n\n\n\n

    Differentiate both:<\/p>\n\n\n\n

      \n
    • u\u2032(x)=ddx(9×4\u2212x+8)=36×3\u22121u'(x) = \\frac{d}{dx}(9x^4 – x + 8) = 36x^3 – 1<\/li>\n\n\n\n
    • v\u2032(x)=ddx(\u2212x5+2)=\u22125x4v'(x) = \\frac{d}{dx}(-x^5 + 2) = -5x^4<\/li>\n<\/ul>\n\n\n\n
      \n\n\n\n

      Step 3: Apply the Product Rule<\/h3>\n\n\n\n

      dydx=u\u2032(x)v(x)+u(x)v\u2032(x)\\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x)<\/p>\n\n\n\n

      Substitute: dydx=(36×3\u22121)(\u2212x5+2)+(9×4\u2212x+8)(\u22125×4)\\frac{dy}{dx} = (36x^3 – 1)(-x^5 + 2) + (9x^4 – x + 8)(-5x^4)<\/p>\n\n\n\n

      \n\n\n\n

      Step 4: Expand both terms<\/h3>\n\n\n\n

      First term:<\/strong> (36×3\u22121)(\u2212x5+2)=36×3(\u2212x5)+36×3(2)\u22121(\u2212x5)\u22121(2)=\u221236×8+72×3+x5\u22122(36x^3 – 1)(-x^5 + 2) = 36x^3(-x^5) + 36x^3(2) -1(-x^5) – 1(2) \\\\ = -36x^8 + 72x^3 + x^5 – 2<\/p>\n\n\n\n

      Second term:<\/strong> (9×4\u2212x+8)(\u22125×4)=9×4(\u22125×4)+(\u2212x)(\u22125×4)+8(\u22125×4)=\u221245×8+5×5\u221240×4(9x^4 – x + 8)(-5x^4) = 9x^4(-5x^4) + (-x)(-5x^4) + 8(-5x^4) \\\\ = -45x^8 + 5x^5 – 40x^4<\/p>\n\n\n\n

      \n\n\n\n

      Step 5: Combine like terms<\/h3>\n\n\n\n

      dydx=(\u221236×8+72×3+x5\u22122)+(\u221245×8+5×5\u221240×4)\\frac{dy}{dx} = (-36x^8 + 72x^3 + x^5 – 2) + (-45x^8 + 5x^5 – 40x^4)<\/p>\n\n\n\n

      Group and combine:<\/p>\n\n\n\n

        \n
      • \u221236×8\u221245×8=\u221281×8-36x^8 – 45x^8 = -81x^8<\/li>\n\n\n\n
      • x5+5×5=6x5x^5 + 5x^5 = 6x^5<\/li>\n\n\n\n
      • \u221240×4-40x^4<\/li>\n\n\n\n
      • +72×3+72x^3<\/li>\n\n\n\n
      • \u22122-2<\/li>\n<\/ul>\n\n\n\n
        \n\n\n\n

        \u2705 Final Answer:<\/h3>\n\n\n\n

        dydx=\u221281×8+6×5\u221240×4+72×3\u22122\\boxed{\\frac{dy}{dx} = -81x^8 + 6x^5 – 40x^4 + 72x^3 – 2}<\/p>\n\n\n\n

        \n\n\n\n

        \ud83d\udd0d Explanation (300 words):<\/h3>\n\n\n\n

        To differentiate a function involving a product of two expressions, the Product Rule<\/strong> is the key tool. The Product Rule allows us to take the derivative of two functions multiplied together without first expanding them. It states that the derivative of y=u(x)v(x)y = u(x)v(x) is u\u2032(x)v(x)+u(x)v\u2032(x)u'(x)v(x) + u(x)v'(x).<\/p>\n\n\n\n

        Here, we define u(x)=9×4\u2212x+8u(x) = 9x^4 – x + 8 and v(x)=\u2212x5+2v(x) = -x^5 + 2. Each part is a polynomial, so we differentiate using basic rules:<\/p>\n\n\n\n

          \n
        • The derivative of 9x49x^4 is 36x336x^3,<\/li>\n\n\n\n
        • The derivative of \u2212x-x is \u22121-1,<\/li>\n\n\n\n
        • Constants like 8 and 2 differentiate to zero.<\/li>\n<\/ul>\n\n\n\n

          Thus, u\u2032(x)=36×3\u22121u'(x) = 36x^3 – 1 and v\u2032(x)=\u22125x4v'(x) = -5x^4.<\/p>\n\n\n\n

          Next, we plug into the Product Rule formula: dydx=u\u2032(x)v(x)+u(x)v\u2032(x)\\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x)<\/p>\n\n\n\n

          We expand both products carefully using the distributive property. After distributing and multiplying terms, we group like terms\u2014specifically powers of xx: x8,x5,x4,x3x^8, x^5, x^4, x^3, and constants. Combining all, we simplify the result.<\/p>\n\n\n\n

          The final answer shows the derivative as a single polynomial, expressed in standard form with descending powers of xx. This systematic approach helps avoid errors and ensures that each component contributes correctly to the derivative.<\/p>\n","protected":false},"excerpt":{"rendered":"

          Differentiate the function y equals left parenthesis 9 x Superscript 4 Baseline minus x plus 8 right parenthesis left parenthesis negative x Superscript 5 Baseline plus 2 right parenthesis The correct answer and explanation is: We are asked to differentiate the function: y=(9×4\u2212x+8)(\u2212x5+2)y = (9x^4 – x + 8)(-x^5 + 2) This is a product […]<\/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-16780","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/16780","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=16780"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/16780\/revisions"}],"predecessor-version":[{"id":16781,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/16780\/revisions\/16781"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=16780"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=16780"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=16780"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}