{"id":42417,"date":"2025-06-29T03:44:04","date_gmt":"2025-06-29T03:44:04","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=42417"},"modified":"2025-06-29T03:44:05","modified_gmt":"2025-06-29T03:44:05","slug":"available-resources","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/available-resources\/","title":{"rendered":"Available Resources"},"content":{"rendered":"\n

Available Resources: \u2022 Desmos Scientific Calculator \u2022 Desmos Matrix Calculator \u2022 Desmos Graphing Calculator \u2022 CalcPlot3D f(x,y)=\\sqrt{2x^2+3y^2} f(2,5) =<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

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

Answer:<\/strong>
The correct answer is 4\/\u221a83<\/strong>.<\/p>\n\n\n\n

Explanation:<\/h3>\n\n\n\n

The problem asks for the value of the partial derivative of the function f(x,y) = \u221a(2x\u00b2 + 3y\u00b2) with respect to x, evaluated at the point (2, 5). This is denoted by f_x(2,5). The process involves two main steps: first finding the partial derivative function f_x(x,y) and then substituting the given values for x and y.<\/p>\n\n\n\n

Step 1: Find the partial derivative f_x(x,y)<\/strong><\/p>\n\n\n\n

To find the partial derivative with respect to x, we treat the variable y as a constant. The function can be rewritten using a fractional exponent, which makes applying differentiation rules easier:
f(x,y) = (2x\u00b2 + 3y\u00b2)^(1\/2)<\/p>\n\n\n\n

We will use the chain rule for differentiation. Let the inner function be u = 2x\u00b2 + 3y\u00b2. The outer function is u^(1\/2).<\/p>\n\n\n\n

    \n
  1. Differentiate the outer function with respect to\u00a0u:
    d\/du (u^(1\/2)) = (1\/2)u^(-1\/2)<\/li>\n\n\n\n
  2. Differentiate the inner function\u00a0u\u00a0with respect to\u00a0x, treating\u00a0y\u00b2\u00a0as a constant:
    \u2202u\/\u2202x = \u2202\/\u2202x (2x\u00b2 + 3y\u00b2) = 4x + 0 = 4x<\/li>\n\n\n\n
  3. Apply the chain rule by multiplying the results from the first two parts:
    f_x(x,y) = (1\/2)(2x\u00b2 + 3y\u00b2)^(-1\/2) * (4x)<\/li>\n<\/ol>\n\n\n\n

    Simplifying this expression gives us:
    f_x(x,y) = (4x) \/ (2\u221a(2x\u00b2 + 3y\u00b2)) = 2x \/ \u221a(2x\u00b2 + 3y\u00b2)<\/p>\n\n\n\n

    Step 2: Evaluate f_x(2,5)<\/strong><\/p>\n\n\n\n

    Now, we substitute x = 2 and y = 5 into the expression for f_x(x,y):
    f_x(2,5) = (2 * 2) \/ \u221a(2 * (2)\u00b2 + 3 * (5)\u00b2)<\/p>\n\n\n\n

    Calculate the numerator:
    2 * 2 = 4<\/p>\n\n\n\n

    Calculate the expression inside the square root in the denominator:
    2 * (2)\u00b2 + 3 * (5)\u00b2 = 2 * 4 + 3 * 25 = 8 + 75 = 83<\/p>\n\n\n\n

    So the expression becomes:
    f_x(2,5) = 4 \/ \u221a83<\/p>\n\n\n\n

    The final value of the partial derivative at the point (2, 5) is 4\/\u221a83<\/strong>.thumb_upthumb_down<\/p>\n\n\n\n

    \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

    Available Resources: \u2022 Desmos Scientific Calculator \u2022 Desmos Matrix Calculator \u2022 Desmos Graphing Calculator \u2022 CalcPlot3D f(x,y)=\\sqrt{2x^2+3y^2} f(2,5) = The Correct Answer and Explanation is: Answer:The correct answer is 4\/\u221a83. Explanation: The problem asks for the value of the partial derivative of the function f(x,y) = \u221a(2x\u00b2 + 3y\u00b2) with respect to x, evaluated at the point (2, 5). […]<\/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-42417","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/42417","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=42417"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/42417\/revisions"}],"predecessor-version":[{"id":42420,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/42417\/revisions\/42420"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=42417"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=42417"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=42417"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}