{"id":185634,"date":"2025-01-22T19:05:39","date_gmt":"2025-01-22T19:05:39","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=185634"},"modified":"2025-01-22T19:05:41","modified_gmt":"2025-01-22T19:05:41","slug":"ditterentiate-x-e-x-1","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/22\/ditterentiate-x-e-x-1\/","title":{"rendered":"Ditterentiate x * e ^ (x + 1)"},"content":{"rendered":"\n

Ditterentiate x * e ^ (x + 1) Compute the 4th derivative of log(x) Compute the derivative of e^(x)\/x at x = 1 Compute the derivative of x^ ^ 2+x^ * y(x) + sin(y(x)) = y(x) . What is the derivative equal to? Compute the derivative of x ^ 2 + y(x) + e ^ (x) = y(x) ^ 2 + 3 What is the derivative equal to? Compute the gradient of f(x,y) = xsin(y). Compute the gradient of f(x,y) = x<\/em>yz-x<\/em>y-1. Integrate log(x) + x Integrate log(x) + x on the interval [1, e] Integrate exp(cos(x ^ 2)) on the interval [1, 3] Find the closed form of the sum of k^3 fork from 0 to n. Compute the limit of e^(-x) as x –> 00. Compute the limit of (cos(x)) \/ (1 – x) as x –> 1. Find the minima and maxima of the function f(x) = x^3+4x^2-7x-5 on the interval [-20, 20]. Find the inflection point of the function f(x) x^3+4<\/em>x^2-7x-5 on the interval [-20, 20] Find the minima and maxima of the function f(x) = xe^(-x) on the interval [-20, 20]. Find the inflection point of the function f(x) = x<\/em>e^(-x) on the interval [-20, 20] What is the equation of the tangent line of f(x) = x*e^(-x) at x = 1? What is the equation of the tangent line of x^ ^ 2+y(x)+3^ * e ^ (x) = y(x) ^ 2 + 3 at x = 0 and y 1? What is the equation of the tangent line of x ^ 2 + y(x) ^ 2 – e ^ (x) = x at x = 0<\/p>\n\n\n\n

The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n

Let’s break this down into manageable parts to answer your questions.<\/p>\n\n\n\n

1. Differentiate ( x \\cdot e^{(x + 1)} )<\/h3>\n\n\n\n

The derivative of ( f(x) = x \\cdot e^{(x+1)} ) can be found using the product rule. The product rule states that the derivative of two functions multiplied together is given by:<\/p>\n\n\n\n

[
\\frac{d}{dx} [f(x) \\cdot g(x)] = f'(x) \\cdot g(x) + f(x) \\cdot g'(x)
]<\/p>\n\n\n\n

For ( f(x) = x ) and ( g(x) = e^{(x+1)} ), we have:<\/p>\n\n\n\n

    \n
  • ( f'(x) = 1 )<\/li>\n\n\n\n
  • ( g'(x) = e^{(x+1)} )<\/li>\n<\/ul>\n\n\n\n

    So, applying the product rule:<\/p>\n\n\n\n

    [
    \\frac{d}{dx}[x \\cdot e^{(x+1)}] = 1 \\cdot e^{(x+1)} + x \\cdot e^{(x+1)} = e^{(x+1)}(1 + x)
    ]<\/p>\n\n\n\n

    2. Compute the 4th derivative of ( \\log(x) )<\/h3>\n\n\n\n

    The derivatives of ( \\log(x) ) are as follows:<\/p>\n\n\n\n

      \n
    • First derivative: ( \\frac{1}{x} )<\/li>\n\n\n\n
    • Second derivative: ( -\\frac{1}{x^2} )<\/li>\n\n\n\n
    • Third derivative: ( \\frac{2}{x^3} )<\/li>\n\n\n\n
    • Fourth derivative: ( -\\frac{6}{x^4} )<\/li>\n<\/ul>\n\n\n\n

      Thus, the fourth derivative of ( \\log(x) ) is ( -\\frac{6}{x^4} ).<\/p>\n\n\n\n

      3. Compute the derivative of ( \\frac{e^x}{x} ) at ( x = 1 )<\/h3>\n\n\n\n

      We can apply the quotient rule to differentiate ( \\frac{e^x}{x} ):
      [
      \\frac{d}{dx} \\left( \\frac{e^x}{x} \\right) = \\frac{x \\cdot e^x – e^x}{x^2} = \\frac{e^x(x – 1)}{x^2}
      ]
      At ( x = 1 ), this becomes:
      [
      \\frac{e^1(1 – 1)}{1^2} = 0
      ]<\/p>\n\n\n\n

      4. Compute the derivative of ( x^2 + x \\cdot y(x) + \\sin(y(x)) = y(x) )<\/h3>\n\n\n\n

      This is an implicit differentiation problem. We differentiate both sides of the equation with respect to ( x ), treating ( y(x) ) as a function of ( x ) and applying the chain rule:<\/p>\n\n\n\n

      [
      \\frac{d}{dx}\\left[ x^2 + x \\cdot y(x) + \\sin(y(x)) \\right] = \\frac{d}{dx}\\left[ y(x) \\right]
      ]<\/p>\n\n\n\n

      This gives:
      [
      2x + \\left( y(x) + x \\cdot y'(x) \\right) + \\cos(y(x)) \\cdot y'(x) = y'(x)
      ]<\/p>\n\n\n\n

      Rearranging this:
      [
      2x + y(x) + y'(x) \\left( x + \\cos(y(x)) – 1 \\right) = 0
      ]
      Thus, the derivative is equal to this expression.<\/p>\n\n\n\n

      5. Compute the derivative of ( x^2 + y(x) + e^x = y(x)^2 + 3 )<\/h3>\n\n\n\n

      Differentiating both sides of the equation:<\/p>\n\n\n\n

      [
      2x + y'(x) + e^x = 2y(x) \\cdot y'(x)
      ]<\/p>\n\n\n\n

      Rearranging for ( y'(x) ):
      [
      y'(x) \\cdot (1 – 2y(x)) = 2x + e^x
      ]
      Thus, the derivative is:
      [
      y'(x) = \\frac{2x + e^x}{2y(x) – 1}
      ]<\/p>\n\n\n\n

      6. Compute the gradient of ( f(x, y) = x \\sin(y) )<\/h3>\n\n\n\n

      The gradient is the vector of partial derivatives with respect to ( x ) and ( y ):
      [
      \\nabla f(x, y) = \\left( \\frac{\\partial}{\\partial x} \\left( x \\sin(y) \\right), \\frac{\\partial}{\\partial y} \\left( x \\sin(y) \\right) \\right)
      ]<\/p>\n\n\n\n

      Thus, we have:
      [
      \\nabla f(x, y) = ( \\sin(y), x \\cos(y) )
      ]<\/p>\n\n\n\n

      7. Compute the gradient of ( f(x, y, z) = x y z – x y – 1 )<\/h3>\n\n\n\n

      The gradient is the vector of partial derivatives with respect to ( x ), ( y ), and ( z ):<\/p>\n\n\n\n

      [
      \\nabla f(x, y, z) = \\left( \\frac{\\partial}{\\partial x} \\left( x y z – x y – 1 \\right), \\frac{\\partial}{\\partial y} \\left( x y z – x y – 1 \\right), \\frac{\\partial}{\\partial z} \\left( x y z – x y – 1 \\right) \\right)
      ]<\/p>\n\n\n\n

      Calculating each derivative:<\/p>\n\n\n\n

        \n
      • With respect to ( x ): ( yz – y )<\/li>\n\n\n\n
      • With respect to ( y ): ( xz – x )<\/li>\n\n\n\n
      • With respect to ( z ): ( xy )<\/li>\n<\/ul>\n\n\n\n

        Thus, the gradient is:
        [
        \\nabla f(x, y, z) = (yz – y, xz – x, xy)
        ]<\/p>\n\n\n\n

        8. Integrate ( \\log(x) + x )<\/h3>\n\n\n\n

        To integrate ( \\log(x) + x ), we break it into two parts:
        [
        \\int \\log(x) \\, dx + \\int x \\, dx
        ]
        The integral of ( \\log(x) ) is ( x \\log(x) – x ), and the integral of ( x ) is ( \\frac{x^2}{2} ), so:
        [
        \\int (\\log(x) + x) \\, dx = x \\log(x) – x + \\frac{x^2}{2} + C
        ]<\/p>\n\n\n\n

        9. Integrate ( \\log(x) + x ) on the interval ( [1, e] )<\/h3>\n\n\n\n

        Substitute the limits into the integral:
        [
        \\left[ x \\log(x) – x + \\frac{x^2}{2} \\right]_{1}^{e}
        ]<\/p>\n\n\n\n

        At ( x = e ):
        [
        e \\log(e) – e + \\frac{e^2}{2} = e – e + \\frac{e^2}{2} = \\frac{e^2}{2}
        ]<\/p>\n\n\n\n

        At ( x = 1 ):
        [
        1 \\log(1) – 1 + \\frac{1^2}{2} = 0 – 1 + \\frac{1}{2} = -\\frac{1}{2}
        ]<\/p>\n\n\n\n

        Thus, the integral is:
        [
        \\frac{e^2}{2} – \\left( -\\frac{1}{2} \\right) = \\frac{e^2}{2} + \\frac{1}{2}
        ]<\/p>\n\n\n\n

        10. Integrate ( \\exp(\\cos(x^2)) ) on the interval ( [1, 3] )<\/h3>\n\n\n\n

        This integral does not have a simple closed-form solution, so it would be solved numerically.<\/p>\n\n\n\n

        11. Find the closed form of the sum of ( k^3 ) for ( k ) from 0 to ( n )<\/h3>\n\n\n\n

        The closed form of the sum of cubes of the first ( n ) integers is:
        [
        \\left( \\frac{n(n+1)}{2} \\right)^2
        ]<\/p>\n\n\n\n

        12. Compute the limit of ( e^{-x} ) as ( x \\to \\infty )<\/h3>\n\n\n\n

        The limit is:
        [
        \\lim_{x \\to \\infty} e^{-x} = 0
        ]<\/p>\n\n\n\n

        13. Compute the limit of ( \\frac{\\cos(x)}{1 – x} ) as ( x \\to 1 )<\/h3>\n\n\n\n

        Substitute ( x = 1 ) into the expression:
        [
        \\lim_{x \\to 1} \\frac{\\cos(x)}{1 – x} = \\frac{\\cos(1)}{0}
        ]
        This limit does not exist as it tends to infinity.<\/p>\n\n\n\n

        14. Find the minima and maxima of the function ( f(x) = x^3 + 4x^2 – 7x – 5 )<\/h3>\n\n\n\n

        We find the critical points by taking the first derivative and setting it to zero:
        [
        f'(x) = 3x^2 + 8x – 7
        ]
        Solving ( 3x^2 + 8x – 7 = 0 ) gives the critical points. Then, we use the second derivative test to identify maxima and minima.<\/p>\n\n\n\n

        15. Find the inflection point of ( f(x) = x^3 + 4x^2 – 7x – 5 )<\/h3>\n\n\n\n

        The inflection point is where the second derivative changes sign. We calculate ( f”(x) ) and solve for where it equals zero.<\/p>\n\n\n\n

        16. Find the minima and maxima of ( f(x) = x \\cdot e^{-x} )<\/h3>\n\n\n\n

        We calculate the first derivative of ( f(x) = x \\cdot e^{-x} ) and set it to zero. Use the second derivative test to find the nature of the critical points.<\/p>\n\n\n\n

        17. Find the inflection point of ( f(x) = x \\cdot e^{-x} )<\/h3>\n\n\n\n

        The second derivative is used to identify the inflection point where the concavity changes.<\/p>\n\n\n\n

        18. Equation of the tangent line of ( f(x) = x \\cdot e^{-x} ) at ( x = 1 )<\/h3>\n\n\n\n

        The slope of the tangent line is given by ( f'(1) ), and the point on the line is ( (1, f(1)) ). The tangent line equation is:
        [
        y – f(1) = f'(1)(x – 1)
        ]<\/p>\n\n\n\n

        19. Equation of the tangent line of ( x^2 + y(x) + 3e^x = y(x)^2 + 3 ) at ( x = 0 ) and ( y(0) = 1 )<\/h3>\n\n\n\n

        We use implicit differentiation to find the slope of the tangent line and then write the equation.<\/p>\n\n\n\n

        20. Equation of the tangent line of ( x^2 + y(x)^2 – e^x = x ) at ( x = 0 )<\/h3>\n\n\n\n

        Again, we use implicit differentiation and solve for the slope and point to write the equation.<\/p>\n\n\n\n

        This explanation covers key steps and highlights for solving these problems. Would you like detailed solutions for any specific part?<\/p>\n","protected":false},"excerpt":{"rendered":"

        Ditterentiate x * e ^ (x + 1) Compute the 4th derivative of log(x) Compute the derivative of e^(x)\/x at x = 1 Compute the derivative of x^ ^ 2+x^ * y(x) + sin(y(x)) = y(x) . What is the derivative equal to? Compute the derivative of x ^ 2 + y(x) + e ^ […]<\/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-185634","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/185634","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=185634"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/185634\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=185634"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=185634"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=185634"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}