{"id":4095,"date":"2022-04-14T18:15:53","date_gmt":"2022-04-14T18:15:53","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/skills-review-for-the-chain-rule\/"},"modified":"2022-11-09T16:37:18","modified_gmt":"2022-11-09T16:37:18","slug":"skills-review-for-the-chain-rule","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/skills-review-for-the-chain-rule\/","title":{"raw":"Skills Review for the Chain Rule","rendered":"Skills Review for the Chain Rule"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Find the derivative of an implicit function by using implicit differentiation<\/li>\r\n \t<li>Write the equation of a line using slope and a point on the line<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn the Chain Rule section, we will learn how to apply the chain rule to functions of several variables. Here we will review implicit differentiation and how to write the equation of a line.\r\n<h2>What is Implicit Differentiation?<\/h2>\r\n<div id=\"fs-id1169737702731\" class=\"bc-section section\">\r\n<p id=\"fs-id1169737766039\"><strong>Implicit differentiation<\/strong> allows us to find slopes of tangents to curves that are clearly not functions (they fail the vertical line test). We are using the idea that portions of [latex]y[\/latex] are functions that satisfy the given equation, but that [latex]y[\/latex] is not actually a function of [latex]x[\/latex].<\/p>\r\n<p id=\"fs-id1169737749590\">To perform implicit differentiation on an equation that defines a function [latex]y[\/latex] implicitly in terms of a variable [latex]x[\/latex], use the following steps:<\/p>\r\n\r\n<ol id=\"fs-id1169737815995\">\r\n \t<li>Take the derivative of both sides of the equation. Keep in mind that [latex]y[\/latex] is a function of [latex]x[\/latex]. Consequently, whereas [latex]\\frac{d}{dx}(\\sin x)= \\cos x, \\, \\frac{d}{dx}(\\sin y)= \\cos y\\frac{dy}{dx}[\/latex] because we must use the Chain Rule to differentiate [latex]\\sin y[\/latex] with respect to [latex]x[\/latex].<\/li>\r\n \t<li>Rewrite the equation so that all terms containing [latex]\\frac{dy}{dx}[\/latex] are on the left and all terms that do not contain [latex]\\frac{dy}{dx}[\/latex] are on the right.<\/li>\r\n \t<li>Factor out [latex]\\frac{dy}{dx}[\/latex] on the left.<\/li>\r\n \t<li>Solve for [latex]\\frac{dy}{dx}[\/latex] by dividing both sides of the equation by an appropriate algebraic expression.<\/li>\r\n<\/ol>\r\n<div id=\"fs-id1169737819953\" class=\"textbook exercises\">\r\n<h3>Example: Using Implicit Differentiation<\/h3>\r\n<p id=\"fs-id1169737931738\">Assuming that [latex]y[\/latex] is defined implicitly by the equation [latex]x^2+y^2=25[\/latex], find [latex]\\frac{dy}{dx}[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1169737948455\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169737948455\"]\r\n<p id=\"fs-id1169737948455\">Follow the steps in the problem-solving strategy.<\/p>\r\n\r\n<div id=\"fs-id1169737840968\" class=\"equation unnumbered\">[latex]\\begin{array}{llll} \\frac{d}{dx}(x^2+y^2) = \\frac{d}{dx}(25) &amp; &amp; &amp; \\text{Step 1. Differentiate both sides of the equation.} \\\\ \\frac{d}{dx}(x^2)+\\frac{d}{dx}(y^2) = 0 &amp; &amp; &amp; \\begin{array}{l}\\text{Step 1.1. Use the sum rule on the left.} \\\\ \\text{On the right,} \\, \\frac{d}{dx}(25)=0. \\end{array} \\\\ 2x+2y\\frac{dy}{dx} = 0 &amp; &amp; &amp; \\begin{array}{l}\\text{Step 1.2. Take the derivatives, so} \\, \\frac{d}{dx}(x^2)=2x \\\\ \\text{and} \\, \\frac{d}{dx}(y^2)=2y\\frac{dy}{dx}. \\end{array} \\\\ 2y\\frac{dy}{dx} = -2x &amp; &amp; &amp; \\begin{array}{l}\\text{Step 2. Keep the terms with} \\, \\frac{dy}{dx} \\, \\text{on the left.} \\\\ \\text{Move the remaining terms to the right.} \\end{array} \\\\ \\frac{dy}{dx} = -\\frac{x}{y} &amp; &amp; &amp; \\begin{array}{l}\\text{Step 4. Divide both sides of the equation by} \\\\ 2y. \\, \\text{(Step 3 does not apply in this case.)} \\end{array} \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<h4>Analysis<\/h4>\r\n<p id=\"fs-id1169737758515\">Note that the resulting expression for [latex]\\frac{dy}{dx}[\/latex] is in terms of both the independent variable [latex]x[\/latex] and the dependent variable [latex]y[\/latex]. Although in some cases it may be possible to express [latex]\\frac{dy}{dx}[\/latex] in terms of [latex]x[\/latex] only, it is generally not possible to do so.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738068274\" class=\"textbook exercises\">\r\n<h3>Example: Using Implicit Differentiation and the Product Rule<\/h3>\r\n<p id=\"fs-id1169737948472\">Assuming that [latex]y[\/latex] is defined implicitly by the equation [latex]x^3 \\sin y+y=4x+3[\/latex], find [latex]\\frac{dy}{dx}[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1169738041617\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169738041617\"]\r\n<div id=\"fs-id1169737814728\" class=\"equation unnumbered\">[latex]\\begin{array}{llll}\\frac{d}{dx}(x^3 \\sin y+y) = \\frac{d}{dx}(4x+3) &amp; &amp; &amp; \\text{Step 1: Differentiate both sides of the equation.} \\\\ \\frac{d}{dx}(x^3 \\sin y)+\\frac{d}{dx}(y) = 4 &amp; &amp; &amp; \\begin{array}{l}\\text{Step 1.1: Apply the sum rule on the left.} \\\\ \\text{On the right,} \\, \\frac{d}{dx}(4x+3)=4. \\end{array} \\\\ (\\frac{d}{dx}(x^3) \\cdot \\sin y+\\frac{d}{dx}(\\sin y) \\cdot x^3) + \\frac{dy}{dx} = 4 &amp; &amp; &amp; \\begin{array}{l}\\text{Step 1.2: Use the product rule to find} \\\\ \\frac{d}{dx}(x^3 \\sin y). \\, \\text{Observe that} \\, \\frac{d}{dx}(y)=\\frac{dy}{dx}. \\end{array} \\\\ 3x^2 \\sin y+(\\cos y\\frac{dy}{dx}) \\cdot x^3 + \\frac{dy}{dx} = 4 &amp; &amp; &amp; \\begin{array}{l}\\text{Step 1.3: We know} \\, \\frac{d}{dx}(x^3)=3x^2. \\, \\text{Use the} \\\\ \\text{chain rule to obtain} \\, \\frac{d}{dx}(\\sin y)= \\cos y\\frac{dy}{dx}. \\end{array} \\\\ x^3 \\cos y\\frac{dy}{dx}+\\frac{dy}{dx} = 4-3x^2 \\sin y &amp; &amp; &amp; \\begin{array}{l}\\text{Step 2: Keep all terms containing} \\, \\frac{dy}{dx} \\, \\text{on the} \\\\ \\text{left. Move all other terms to the right.} \\end{array} \\\\ \\frac{dy}{dx}(x^3 \\cos y+1) = 4-3x^2 \\sin y &amp; &amp; &amp; \\text{Step 3: Factor out} \\, \\frac{dy}{dx} \\, \\text{on the left.} \\\\ \\frac{dy}{dx} = \\large \\frac{4-3x^2 \\sin y}{x^3 \\cos y+1} &amp; &amp; &amp; \\begin{array}{l}\\text{Step 4: Solve for} \\, \\frac{dy}{dx} \\, \\text{by dividing both sides of} \\\\ \\text{the equation by} \\, x^3 \\cos y+1. \\end{array} \\end{array}[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169737772805\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1169738220225\">Find [latex]\\frac{dy}{dx}[\/latex] for [latex]y[\/latex] defined implicitly by the equation [latex]4x^5+ \\tan y=y^2+5x[\/latex].<\/p>\r\n[reveal-answer q=\"8993550\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"8993550\"]\r\n<p id=\"fs-id1169738221942\">Follow the problem solving strategy, remembering to apply the chain rule to differentiate [latex]\\tan y[\/latex] and [latex]y^2[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1169737953797\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169737953797\"]\r\n<p id=\"fs-id1169737953797\">[latex]\\frac{dy}{dx}=\\large \\frac{5-20x^4}{\\sec^2 y-2y}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]206109[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Write the Equation of a Line<\/h2>\r\n<strong><em>(also in Module 1, Skills Review for Parametric Equations and Calculus of Parametric Equations)<\/em><\/strong>\r\nThe <strong>slope<\/strong> of a line refers to the ratio of the vertical change in <em>y<\/em> over the horizontal change in <em>x<\/em> between any two points on a line.\r\n<div class=\"textbox\">\r\n<h3>A General Note: The Slope of a Line<\/h3>\r\nThe slope of a line, <em>m<\/em>, represents the change in <em>y<\/em> over the change in <em>x.<\/em> Given two points, [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex] and [latex]\\left({x}_{2},{y}_{2}\\right)[\/latex], the following formula determines the slope of a line containing these points:\r\n<div style=\"text-align: center;\">[latex]m=\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}[\/latex]<\/div>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the Slope of a Line Given Two Points<\/h3>\r\nFind the slope of a line that passes through the points [latex]\\left(2,-1\\right)[\/latex] and [latex]\\left(-5,3\\right)[\/latex].\r\n[reveal-answer q=\"688301\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"688301\"]\r\n\r\nWe substitute the <em>y-<\/em>values and the <em>x-<\/em>values into the formula.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}m\\hfill&amp;=\\frac{3-\\left(-1\\right)}{-5 - 2}\\hfill \\\\ \\hfill&amp;=\\frac{4}{-7}\\hfill \\\\ \\hfill&amp;=-\\frac{4}{7}\\hfill \\end{array}[\/latex]<\/div>\r\nThe slope is [latex]-\\frac{4}{7}[\/latex].\r\n<h4>Analysis of the Solution<\/h4>\r\nIt does not matter which point is called [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex] or [latex]\\left({x}_{2},{y}_{2}\\right)[\/latex]. As long as we are consistent with the order of the <em>y<\/em> terms and the order of the <em>x<\/em> terms in the numerator and denominator, the calculation will yield the same result.\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFind the slope of the line that passes through the points [latex]\\left(-2,6\\right)[\/latex] and [latex]\\left(1,4\\right)[\/latex].\r\n\r\n[reveal-answer q=\"196055\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"196055\"]\r\n\r\nslope[latex]=m=\\dfrac{-2}{3}=-\\dfrac{2}{3}[\/latex][\/hidden-answer]\r\n\r\n[ohm_question]1719[\/ohm_question]\r\n\r\n<\/div>\r\nTo write the equation of a line, the line's slope and a point the line goes through must be known. Perhaps the most familiar form of a linear equation is <strong>slope-intercept form<\/strong> written as [latex]y=mx+b[\/latex], where [latex]m=\\text{slope}[\/latex] and [latex]b=y\\text{-intercept}[\/latex]. Let us begin with the slope.\r\n\r\nOften, the starting point to writing the equation of a line is to use <strong>point-slope formula<\/strong>.\u00a0Given the slope and one point on a line, we can find the equation of the line using point-slope form shown below.\r\n<div style=\"text-align: center;\">[latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/div>\r\nWe need only one point and the slope of the line to use the formula. After substituting the slope and the coordinates of one point into the formula, we simplify it and write it in slope-intercept form.\r\n<div class=\"textbox\">\r\n<h3>A General Note: The Point-Slope Formula<\/h3>\r\nGiven one point and the slope, using point-slope form will lead to the equation of a line:\r\n<div style=\"text-align: center;\">[latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/div>\r\n<\/div>\r\n<div style=\"text-align: left;\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the Equation of a Line Given the Slope and One Point<\/h3>\r\nWrite the equation of the line with slope [latex]m=-3[\/latex] and passing through the point [latex]\\left(4,8\\right)[\/latex]. Write the final equation in slope-intercept form.\r\n[reveal-answer q=\"201330\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"201330\"]\r\n\r\nUsing point-slope form, substitute [latex]-3[\/latex] for <em>m <\/em>and the point [latex]\\left(4,8\\right)[\/latex] for [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex].\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\hfill \\\\ y - 8=-3\\left(x - 4\\right)\\hfill \\\\ y - 8=-3x+12\\hfill \\\\ y=-3x+20\\hfill \\end{array}[\/latex]<\/div>\r\n<div>\r\n<div>\r\n<h4>Analysis of the Solution<\/h4>\r\n<\/div>\r\n<div>\r\n\r\nNote that any point on the line can be used to find the equation. If done correctly, the same final equation will be obtained.\r\n\r\n<\/div>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nGiven [latex]m=4[\/latex], find the equation of the line in slope-intercept form passing through the point [latex]\\left(2,5\\right)[\/latex].\r\n\r\n[reveal-answer q=\"634647\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"634647\"]\r\n\r\n[latex]y=4x - 3[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]110942[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Find the derivative of an implicit function by using implicit differentiation<\/li>\n<li>Write the equation of a line using slope and a point on the line<\/li>\n<\/ul>\n<\/div>\n<p>In the Chain Rule section, we will learn how to apply the chain rule to functions of several variables. Here we will review implicit differentiation and how to write the equation of a line.<\/p>\n<h2>What is Implicit Differentiation?<\/h2>\n<div id=\"fs-id1169737702731\" class=\"bc-section section\">\n<p id=\"fs-id1169737766039\"><strong>Implicit differentiation<\/strong> allows us to find slopes of tangents to curves that are clearly not functions (they fail the vertical line test). We are using the idea that portions of [latex]y[\/latex] are functions that satisfy the given equation, but that [latex]y[\/latex] is not actually a function of [latex]x[\/latex].<\/p>\n<p id=\"fs-id1169737749590\">To perform implicit differentiation on an equation that defines a function [latex]y[\/latex] implicitly in terms of a variable [latex]x[\/latex], use the following steps:<\/p>\n<ol id=\"fs-id1169737815995\">\n<li>Take the derivative of both sides of the equation. Keep in mind that [latex]y[\/latex] is a function of [latex]x[\/latex]. Consequently, whereas [latex]\\frac{d}{dx}(\\sin x)= \\cos x, \\, \\frac{d}{dx}(\\sin y)= \\cos y\\frac{dy}{dx}[\/latex] because we must use the Chain Rule to differentiate [latex]\\sin y[\/latex] with respect to [latex]x[\/latex].<\/li>\n<li>Rewrite the equation so that all terms containing [latex]\\frac{dy}{dx}[\/latex] are on the left and all terms that do not contain [latex]\\frac{dy}{dx}[\/latex] are on the right.<\/li>\n<li>Factor out [latex]\\frac{dy}{dx}[\/latex] on the left.<\/li>\n<li>Solve for [latex]\\frac{dy}{dx}[\/latex] by dividing both sides of the equation by an appropriate algebraic expression.<\/li>\n<\/ol>\n<div id=\"fs-id1169737819953\" class=\"textbook exercises\">\n<h3>Example: Using Implicit Differentiation<\/h3>\n<p id=\"fs-id1169737931738\">Assuming that [latex]y[\/latex] is defined implicitly by the equation [latex]x^2+y^2=25[\/latex], find [latex]\\frac{dy}{dx}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169737948455\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169737948455\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169737948455\">Follow the steps in the problem-solving strategy.<\/p>\n<div id=\"fs-id1169737840968\" class=\"equation unnumbered\">[latex]\\begin{array}{llll} \\frac{d}{dx}(x^2+y^2) = \\frac{d}{dx}(25) & & & \\text{Step 1. Differentiate both sides of the equation.} \\\\ \\frac{d}{dx}(x^2)+\\frac{d}{dx}(y^2) = 0 & & & \\begin{array}{l}\\text{Step 1.1. Use the sum rule on the left.} \\\\ \\text{On the right,} \\, \\frac{d}{dx}(25)=0. \\end{array} \\\\ 2x+2y\\frac{dy}{dx} = 0 & & & \\begin{array}{l}\\text{Step 1.2. Take the derivatives, so} \\, \\frac{d}{dx}(x^2)=2x \\\\ \\text{and} \\, \\frac{d}{dx}(y^2)=2y\\frac{dy}{dx}. \\end{array} \\\\ 2y\\frac{dy}{dx} = -2x & & & \\begin{array}{l}\\text{Step 2. Keep the terms with} \\, \\frac{dy}{dx} \\, \\text{on the left.} \\\\ \\text{Move the remaining terms to the right.} \\end{array} \\\\ \\frac{dy}{dx} = -\\frac{x}{y} & & & \\begin{array}{l}\\text{Step 4. Divide both sides of the equation by} \\\\ 2y. \\, \\text{(Step 3 does not apply in this case.)} \\end{array} \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<h4>Analysis<\/h4>\n<p id=\"fs-id1169737758515\">Note that the resulting expression for [latex]\\frac{dy}{dx}[\/latex] is in terms of both the independent variable [latex]x[\/latex] and the dependent variable [latex]y[\/latex]. Although in some cases it may be possible to express [latex]\\frac{dy}{dx}[\/latex] in terms of [latex]x[\/latex] only, it is generally not possible to do so.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738068274\" class=\"textbook exercises\">\n<h3>Example: Using Implicit Differentiation and the Product Rule<\/h3>\n<p id=\"fs-id1169737948472\">Assuming that [latex]y[\/latex] is defined implicitly by the equation [latex]x^3 \\sin y+y=4x+3[\/latex], find [latex]\\frac{dy}{dx}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169738041617\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169738041617\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169737814728\" class=\"equation unnumbered\">[latex]\\begin{array}{llll}\\frac{d}{dx}(x^3 \\sin y+y) = \\frac{d}{dx}(4x+3) & & & \\text{Step 1: Differentiate both sides of the equation.} \\\\ \\frac{d}{dx}(x^3 \\sin y)+\\frac{d}{dx}(y) = 4 & & & \\begin{array}{l}\\text{Step 1.1: Apply the sum rule on the left.} \\\\ \\text{On the right,} \\, \\frac{d}{dx}(4x+3)=4. \\end{array} \\\\ (\\frac{d}{dx}(x^3) \\cdot \\sin y+\\frac{d}{dx}(\\sin y) \\cdot x^3) + \\frac{dy}{dx} = 4 & & & \\begin{array}{l}\\text{Step 1.2: Use the product rule to find} \\\\ \\frac{d}{dx}(x^3 \\sin y). \\, \\text{Observe that} \\, \\frac{d}{dx}(y)=\\frac{dy}{dx}. \\end{array} \\\\ 3x^2 \\sin y+(\\cos y\\frac{dy}{dx}) \\cdot x^3 + \\frac{dy}{dx} = 4 & & & \\begin{array}{l}\\text{Step 1.3: We know} \\, \\frac{d}{dx}(x^3)=3x^2. \\, \\text{Use the} \\\\ \\text{chain rule to obtain} \\, \\frac{d}{dx}(\\sin y)= \\cos y\\frac{dy}{dx}. \\end{array} \\\\ x^3 \\cos y\\frac{dy}{dx}+\\frac{dy}{dx} = 4-3x^2 \\sin y & & & \\begin{array}{l}\\text{Step 2: Keep all terms containing} \\, \\frac{dy}{dx} \\, \\text{on the} \\\\ \\text{left. Move all other terms to the right.} \\end{array} \\\\ \\frac{dy}{dx}(x^3 \\cos y+1) = 4-3x^2 \\sin y & & & \\text{Step 3: Factor out} \\, \\frac{dy}{dx} \\, \\text{on the left.} \\\\ \\frac{dy}{dx} = \\large \\frac{4-3x^2 \\sin y}{x^3 \\cos y+1} & & & \\begin{array}{l}\\text{Step 4: Solve for} \\, \\frac{dy}{dx} \\, \\text{by dividing both sides of} \\\\ \\text{the equation by} \\, x^3 \\cos y+1. \\end{array} \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169737772805\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1169738220225\">Find [latex]\\frac{dy}{dx}[\/latex] for [latex]y[\/latex] defined implicitly by the equation [latex]4x^5+ \\tan y=y^2+5x[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q8993550\">Hint<\/span><\/p>\n<div id=\"q8993550\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738221942\">Follow the problem solving strategy, remembering to apply the chain rule to differentiate [latex]\\tan y[\/latex] and [latex]y^2[\/latex].<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169737953797\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169737953797\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169737953797\">[latex]\\frac{dy}{dx}=\\large \\frac{5-20x^4}{\\sec^2 y-2y}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm206109\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=206109&theme=oea&iframe_resize_id=ohm206109&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Write the Equation of a Line<\/h2>\n<p><strong><em>(also in Module 1, Skills Review for Parametric Equations and Calculus of Parametric Equations)<\/em><\/strong><br \/>\nThe <strong>slope<\/strong> of a line refers to the ratio of the vertical change in <em>y<\/em> over the horizontal change in <em>x<\/em> between any two points on a line.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: The Slope of a Line<\/h3>\n<p>The slope of a line, <em>m<\/em>, represents the change in <em>y<\/em> over the change in <em>x.<\/em> Given two points, [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex] and [latex]\\left({x}_{2},{y}_{2}\\right)[\/latex], the following formula determines the slope of a line containing these points:<\/p>\n<div style=\"text-align: center;\">[latex]m=\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}[\/latex]<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Slope of a Line Given Two Points<\/h3>\n<p>Find the slope of a line that passes through the points [latex]\\left(2,-1\\right)[\/latex] and [latex]\\left(-5,3\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q688301\">Show Solution<\/span><\/p>\n<div id=\"q688301\" class=\"hidden-answer\" style=\"display: none\">\n<p>We substitute the <em>y-<\/em>values and the <em>x-<\/em>values into the formula.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}m\\hfill&=\\frac{3-\\left(-1\\right)}{-5 - 2}\\hfill \\\\ \\hfill&=\\frac{4}{-7}\\hfill \\\\ \\hfill&=-\\frac{4}{7}\\hfill \\end{array}[\/latex]<\/div>\n<p>The slope is [latex]-\\frac{4}{7}[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>It does not matter which point is called [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex] or [latex]\\left({x}_{2},{y}_{2}\\right)[\/latex]. As long as we are consistent with the order of the <em>y<\/em> terms and the order of the <em>x<\/em> terms in the numerator and denominator, the calculation will yield the same result.\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Find the slope of the line that passes through the points [latex]\\left(-2,6\\right)[\/latex] and [latex]\\left(1,4\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q196055\">Show Solution<\/span><\/p>\n<div id=\"q196055\" class=\"hidden-answer\" style=\"display: none\">\n<p>slope[latex]=m=\\dfrac{-2}{3}=-\\dfrac{2}{3}[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"ohm1719\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=1719&theme=oea&iframe_resize_id=ohm1719&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>To write the equation of a line, the line&#8217;s slope and a point the line goes through must be known. Perhaps the most familiar form of a linear equation is <strong>slope-intercept form<\/strong> written as [latex]y=mx+b[\/latex], where [latex]m=\\text{slope}[\/latex] and [latex]b=y\\text{-intercept}[\/latex]. Let us begin with the slope.<\/p>\n<p>Often, the starting point to writing the equation of a line is to use <strong>point-slope formula<\/strong>.\u00a0Given the slope and one point on a line, we can find the equation of the line using point-slope form shown below.<\/p>\n<div style=\"text-align: center;\">[latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/div>\n<p>We need only one point and the slope of the line to use the formula. After substituting the slope and the coordinates of one point into the formula, we simplify it and write it in slope-intercept form.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: The Point-Slope Formula<\/h3>\n<p>Given one point and the slope, using point-slope form will lead to the equation of a line:<\/p>\n<div style=\"text-align: center;\">[latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/div>\n<\/div>\n<div style=\"text-align: left;\">\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Equation of a Line Given the Slope and One Point<\/h3>\n<p>Write the equation of the line with slope [latex]m=-3[\/latex] and passing through the point [latex]\\left(4,8\\right)[\/latex]. Write the final equation in slope-intercept form.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q201330\">Show Solution<\/span><\/p>\n<div id=\"q201330\" class=\"hidden-answer\" style=\"display: none\">\n<p>Using point-slope form, substitute [latex]-3[\/latex] for <em>m <\/em>and the point [latex]\\left(4,8\\right)[\/latex] for [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\hfill \\\\ y - 8=-3\\left(x - 4\\right)\\hfill \\\\ y - 8=-3x+12\\hfill \\\\ y=-3x+20\\hfill \\end{array}[\/latex]<\/div>\n<div>\n<div>\n<h4>Analysis of the Solution<\/h4>\n<\/div>\n<div>\n<p>Note that any point on the line can be used to find the equation. If done correctly, the same final equation will be obtained.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Given [latex]m=4[\/latex], find the equation of the line in slope-intercept form passing through the point [latex]\\left(2,5\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q634647\">Show Solution<\/span><\/p>\n<div id=\"q634647\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]y=4x - 3[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm110942\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=110942&theme=oea&iframe_resize_id=ohm110942&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-4095\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 1. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/calculus1\/\">https:\/\/courses.lumenlearning.com\/calculus1\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Calculus Volume 2. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/calculus2\/\">https:\/\/courses.lumenlearning.com\/calculus2\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":349141,"menu_order":3,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 1\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"https:\/\/courses.lumenlearning.com\/calculus1\/\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Calculus Volume 2\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"https:\/\/courses.lumenlearning.com\/calculus2\/\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-4095","chapter","type-chapter","status-publish","hentry"],"part":4126,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/4095","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/349141"}],"version-history":[{"count":8,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/4095\/revisions"}],"predecessor-version":[{"id":6484,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/4095\/revisions\/6484"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/4126"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/4095\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=4095"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=4095"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=4095"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=4095"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}