{"id":261,"date":"2023-09-20T22:48:39","date_gmt":"2023-09-20T22:48:39","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/finding-the-derivative-using-implicit-differentiation\/"},"modified":"2024-08-05T12:54:58","modified_gmt":"2024-08-05T12:54:58","slug":"implicit-differentiation-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/implicit-differentiation-learn-it-1\/","title":{"raw":"Implicit Differentiation: Learn It 1","rendered":"Implicit Differentiation: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\">\r\n<ul>\r\n\t<li>Use implicit differentiation to find derivatives and the equations for tangent lines<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>What is Implicit Differentiation?<\/h2>\r\n<p>We have previously explored how to find tangent lines to functions and determine the rate of change of a function at a specific point by explicitly defining the function and its derivatives. However, not all functions can be expressed directly in terms of one variable. Implicit differentiation comes into play when we need to derive functions that are defined implicitly rather than explicitly.\u00a0<\/p>\r\n<section class=\"textbox recall\">\r\n<p>A function is typically described as explicit when the dependent variable [latex]y[\/latex] is expressed solely in terms of the independent variable [latex]x[\/latex]. For example, [latex]y=x^2+1[\/latex] is an <strong>explicit function<\/strong> because [latex]y[\/latex] is given directly in terms of [latex]x[\/latex].<\/p>\r\n<p>Conversely, if [latex]y[\/latex] and [latex]x[\/latex] are interrelated through an equation without a straightforward expression of [latex]y[\/latex] in terms of [latex]x[\/latex], the function is considered an <strong>implicit function<\/strong>. An example is the circle equation [latex]x^2 +y^2 =25[\/latex], which doesn\u2019t solve for [latex]y[\/latex] explicitly in terms of [latex]x[\/latex].<\/p>\r\n<\/section>\r\n<p id=\"fs-id1169737766039\"><strong>Implicit differentiation<\/strong> is essential for finding the slopes of tangent lines to curves that are not explicit functions. For example, the equation [latex]y-x^2=1[\/latex] implicitly defines [latex]y[\/latex] because it does not isolate [latex]y[\/latex] on one side of the equation. This method allows us to derive relations where [latex]y[\/latex] is defined implicitly by differentiating both sides of the equation with respect to [latex]x[\/latex], treating [latex]y[\/latex] as a function of [latex]x[\/latex] where necessary.<\/p>\r\n<p id=\"fs-id1169738018789\">An equation defines a function implicitly if it holds for [latex]y[\/latex] and [latex]x[\/latex] without isolating one variable on one side. For instance, the equations:<\/p>\r\n<p id=\"fs-id1169737771295\" style=\"text-align: center;\">[latex]y=\\sqrt{25-x^2}[\/latex],\u00a0 [latex]y = -\\sqrt{25-x^2}[\/latex], and [latex]y=\\begin{cases} \\sqrt{25-x^2} &amp; \\text{ if } \\, -5 \\le x &lt; 0 \\\\ -\\sqrt{25-x^2} &amp; \\text{ if } \\, 0 \\le x \\le 5 \\end{cases}[\/latex],<\/p>\r\n<p>are all examples of functions defined implicitly by the circle equation [latex]x^2+y^2=25[\/latex].<\/p>\r\n<p>Figure 1 illustrates how the equation [latex]x^2+y^2=25[\/latex] defines multiple functions implicitly, showcasing different ways [latex]y[\/latex] can be expressed in relation to [latex]x[\/latex]. These representations include the full circle and its segments, dependent on [latex]x[\/latex] values, demonstrating the versatility of implicit functions.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"875\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11205502\/CNX_Calc_Figure_03_08_001.jpg\" alt=\"The circle with radius 5 and center at the origin is graphed fully in one picture. Then, only its segments in quadrants I and II are graphed. Then, only its segments in quadrants III and IV are graphed. Lastly, only its segments in quadrants II and IV are graphed.\" width=\"875\" height=\"988\" \/> Figure 1. The equation [latex]{x}^{2}+{y}^{2}=25[\/latex] defines many functions implicitly.[\/caption]\r\n\r\n<p id=\"fs-id1169737773822\">To apply implicit differentiation practically, consider finding the slope of the tangent line to the circle at a specific point.<\/p>\r\n<p>For example, to determine the slope at point [latex](3,4)[\/latex], one might initially think to differentiate [latex]y=\\sqrt{25-x^2}[\/latex] directly at [latex]x=3[\/latex]. Similarly, to find the slope at [latex](3,-4)[\/latex], we could use the derivative of [latex]y=\u2212\\sqrt{25-x^2}[\/latex]. However, these expressions don't always provide the clearest path for differentiation, especially over different ranges of [latex]x[\/latex].<\/p>\r\n<p>Implicit differentiation streamlines this process by differentiating the entire equation directly:\u00a0<\/p>\r\n<p style=\"text-align: center;\">[latex]2x+2y \\frac{dy}{dx}=0[\/latex]<\/p>\r\n<p>This leads to:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\frac{dy}{dx}=\\frac{-x}{y}[\/latex]<\/p>\r\n<p>which can then be evaluated at any point on the curve without having to solve explicitly for [latex]y[\/latex] first. This method confirms the flexibility and power of implicit differentiation in handling equations where [latex]y[\/latex] is not isolated.<\/p>\r\n<p>The process of finding [latex]\\frac{dy}{dx}[\/latex] using implicit differentiation is described in the following problem-solving strategy.<\/p>\r\n<section class=\"textbox questionHelp\">\r\n<p><strong>Problem-Solving Strategy: Implicit Differentiation<\/strong><\/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<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<\/section>\r\n<section class=\"textbox example\">\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<p>[reveal-answer q=\"fs-id1169737948455\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1169737948455\"]<\/p>\r\n<p id=\"fs-id1169737948455\">Follow the steps in the problem-solving strategy.<\/p>\r\n<div id=\"fs-id1169737840968\" class=\"equation unnumbered\" style=\"text-align: center;\">[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<p><strong>Analysis<\/strong><\/p>\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<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">\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<p>[reveal-answer q=\"fs-id1169738041617\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1169738041617\"]<\/p>\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<\/section>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1169738211084\">Find [latex]\\frac{d^2 y}{dx^2}[\/latex] if [latex]x^2+y^2=25[\/latex].<\/p>\r\n<p>[reveal-answer q=\"fs-id1169738226743\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1169738226743\"]<\/p>\r\n<p id=\"fs-id1169738226743\">In the first example, we showed that [latex]\\frac{dy}{dx}=-\\frac{x}{y}[\/latex]. We can take the derivative of both sides of this equation to find [latex]\\frac{d^2 y}{dx^2}[\/latex].<\/p>\r\n<div id=\"fs-id1169737766542\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lllll} \\frac{d^2 y}{dx^2} &amp; =\\large \\frac{d}{dx}(-\\frac{x}{y}) &amp; &amp; &amp; \\text{Differentiate both sides of} \\, \\frac{dy}{dx}=-\\frac{x}{y}. \\\\ &amp; = \\large -\\frac{(1 \\cdot y-x\\frac{dy}{dx})}{y^2} &amp; &amp; &amp; \\text{Use the quotient rule to find} \\, \\frac{d}{dx}(-\\frac{x}{y}). \\\\ &amp; = \\large \\frac{\u2212y+x\\frac{dy}{dx}}{y^2} &amp; &amp; &amp; \\text{Simplify.} \\\\ &amp; = \\large \\frac{\u2212y+x(-\\frac{x}{y})}{y^2} &amp; &amp; &amp; \\text{Substitute} \\, \\frac{dy}{dx}=-\\frac{x}{y}. \\\\ &amp; = \\large \\frac{\u2212y^2-x^2}{y^3} &amp; &amp; &amp; \\text{Simplify.} \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1169738044967\">At this point we have found an expression for [latex]\\frac{d^2 y}{dx^2}[\/latex]. If we choose, we can simplify the expression further by recalling that [latex]x^2+y^2=25[\/latex] and making this substitution in the numerator to obtain [latex]\\frac{d^2 y}{dx^2}=-\\frac{25}{y^3}[\/latex].<\/p>\r\n<p>Watch the following video to see the worked solution to this example..<\/p>\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/F1FFAg-XhlQ?controls=0&amp;start=385&amp;end=473&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>\r\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\r\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/3.8ImplicitDifferentiation385to473_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"3.8 Implicit Differentiation\" here (opens in new window)<\/a>.<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox tryIt\">\r\n<p>[ohm_question hide_question_numbers=1]206109[\/ohm_question]<\/p>\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Use implicit differentiation to find derivatives and the equations for tangent lines<\/li>\n<\/ul>\n<\/section>\n<h2>What is Implicit Differentiation?<\/h2>\n<p>We have previously explored how to find tangent lines to functions and determine the rate of change of a function at a specific point by explicitly defining the function and its derivatives. However, not all functions can be expressed directly in terms of one variable. Implicit differentiation comes into play when we need to derive functions that are defined implicitly rather than explicitly.\u00a0<\/p>\n<section class=\"textbox recall\">\n<p>A function is typically described as explicit when the dependent variable [latex]y[\/latex] is expressed solely in terms of the independent variable [latex]x[\/latex]. For example, [latex]y=x^2+1[\/latex] is an <strong>explicit function<\/strong> because [latex]y[\/latex] is given directly in terms of [latex]x[\/latex].<\/p>\n<p>Conversely, if [latex]y[\/latex] and [latex]x[\/latex] are interrelated through an equation without a straightforward expression of [latex]y[\/latex] in terms of [latex]x[\/latex], the function is considered an <strong>implicit function<\/strong>. An example is the circle equation [latex]x^2 +y^2 =25[\/latex], which doesn\u2019t solve for [latex]y[\/latex] explicitly in terms of [latex]x[\/latex].<\/p>\n<\/section>\n<p id=\"fs-id1169737766039\"><strong>Implicit differentiation<\/strong> is essential for finding the slopes of tangent lines to curves that are not explicit functions. For example, the equation [latex]y-x^2=1[\/latex] implicitly defines [latex]y[\/latex] because it does not isolate [latex]y[\/latex] on one side of the equation. This method allows us to derive relations where [latex]y[\/latex] is defined implicitly by differentiating both sides of the equation with respect to [latex]x[\/latex], treating [latex]y[\/latex] as a function of [latex]x[\/latex] where necessary.<\/p>\n<p id=\"fs-id1169738018789\">An equation defines a function implicitly if it holds for [latex]y[\/latex] and [latex]x[\/latex] without isolating one variable on one side. For instance, the equations:<\/p>\n<p id=\"fs-id1169737771295\" style=\"text-align: center;\">[latex]y=\\sqrt{25-x^2}[\/latex],\u00a0 [latex]y = -\\sqrt{25-x^2}[\/latex], and [latex]y=\\begin{cases} \\sqrt{25-x^2} & \\text{ if } \\, -5 \\le x < 0 \\\\ -\\sqrt{25-x^2} & \\text{ if } \\, 0 \\le x \\le 5 \\end{cases}[\/latex],<\/p>\n<p>are all examples of functions defined implicitly by the circle equation [latex]x^2+y^2=25[\/latex].<\/p>\n<p>Figure 1 illustrates how the equation [latex]x^2+y^2=25[\/latex] defines multiple functions implicitly, showcasing different ways [latex]y[\/latex] can be expressed in relation to [latex]x[\/latex]. These representations include the full circle and its segments, dependent on [latex]x[\/latex] values, demonstrating the versatility of implicit functions.<\/p>\n<figure style=\"width: 875px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11205502\/CNX_Calc_Figure_03_08_001.jpg\" alt=\"The circle with radius 5 and center at the origin is graphed fully in one picture. Then, only its segments in quadrants I and II are graphed. Then, only its segments in quadrants III and IV are graphed. Lastly, only its segments in quadrants II and IV are graphed.\" width=\"875\" height=\"988\" \/><figcaption class=\"wp-caption-text\">Figure 1. The equation [latex]{x}^{2}+{y}^{2}=25[\/latex] defines many functions implicitly.<\/figcaption><\/figure>\n<p id=\"fs-id1169737773822\">To apply implicit differentiation practically, consider finding the slope of the tangent line to the circle at a specific point.<\/p>\n<p>For example, to determine the slope at point [latex](3,4)[\/latex], one might initially think to differentiate [latex]y=\\sqrt{25-x^2}[\/latex] directly at [latex]x=3[\/latex]. Similarly, to find the slope at [latex](3,-4)[\/latex], we could use the derivative of [latex]y=\u2212\\sqrt{25-x^2}[\/latex]. However, these expressions don&#8217;t always provide the clearest path for differentiation, especially over different ranges of [latex]x[\/latex].<\/p>\n<p>Implicit differentiation streamlines this process by differentiating the entire equation directly:\u00a0<\/p>\n<p style=\"text-align: center;\">[latex]2x+2y \\frac{dy}{dx}=0[\/latex]<\/p>\n<p>This leads to:<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{dy}{dx}=\\frac{-x}{y}[\/latex]<\/p>\n<p>which can then be evaluated at any point on the curve without having to solve explicitly for [latex]y[\/latex] first. This method confirms the flexibility and power of implicit differentiation in handling equations where [latex]y[\/latex] is not isolated.<\/p>\n<p>The process of finding [latex]\\frac{dy}{dx}[\/latex] using implicit differentiation is described in the following problem-solving strategy.<\/p>\n<section class=\"textbox questionHelp\">\n<p><strong>Problem-Solving Strategy: Implicit Differentiation<\/strong><\/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<\/section>\n<section class=\"textbox example\">\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<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1169737948455\">Show Solution<\/button><\/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\" style=\"text-align: center;\">[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><strong>Analysis<\/strong><\/p>\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<\/section>\n<section class=\"textbox example\">\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<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1169738041617\">Show Solution<\/button><\/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<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1169738211084\">Find [latex]\\frac{d^2 y}{dx^2}[\/latex] if [latex]x^2+y^2=25[\/latex].<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1169738226743\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1169738226743\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738226743\">In the first example, we showed that [latex]\\frac{dy}{dx}=-\\frac{x}{y}[\/latex]. We can take the derivative of both sides of this equation to find [latex]\\frac{d^2 y}{dx^2}[\/latex].<\/p>\n<div id=\"fs-id1169737766542\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lllll} \\frac{d^2 y}{dx^2} & =\\large \\frac{d}{dx}(-\\frac{x}{y}) & & & \\text{Differentiate both sides of} \\, \\frac{dy}{dx}=-\\frac{x}{y}. \\\\ & = \\large -\\frac{(1 \\cdot y-x\\frac{dy}{dx})}{y^2} & & & \\text{Use the quotient rule to find} \\, \\frac{d}{dx}(-\\frac{x}{y}). \\\\ & = \\large \\frac{\u2212y+x\\frac{dy}{dx}}{y^2} & & & \\text{Simplify.} \\\\ & = \\large \\frac{\u2212y+x(-\\frac{x}{y})}{y^2} & & & \\text{Substitute} \\, \\frac{dy}{dx}=-\\frac{x}{y}. \\\\ & = \\large \\frac{\u2212y^2-x^2}{y^3} & & & \\text{Simplify.} \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1169738044967\">At this point we have found an expression for [latex]\\frac{d^2 y}{dx^2}[\/latex]. If we choose, we can simplify the expression further by recalling that [latex]x^2+y^2=25[\/latex] and making this substitution in the numerator to obtain [latex]\\frac{d^2 y}{dx^2}=-\\frac{25}{y^3}[\/latex].<\/p>\n<p>Watch the following video to see the worked solution to this example..<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/F1FFAg-XhlQ?controls=0&amp;start=385&amp;end=473&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/3.8ImplicitDifferentiation385to473_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;3.8 Implicit Differentiation&#8221; here (opens in new window)<\/a>.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\">\n<iframe loading=\"lazy\" id=\"ohm206109\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=206109&theme=lumen&iframe_resize_id=ohm206109&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><br \/>\n<\/section>\n","protected":false},"author":6,"menu_order":19,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 1\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"3.8 Implicit Differentiation\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":560,"module-header":"learn_it","content_attributions":[{"type":"cc","description":"Calculus Volume 1","author":"Gilbert Strang, Edwin (Jed) Herman","organization":"OpenStax","url":"https:\/\/openstax.org\/details\/books\/calculus-volume-1","project":"","license":"cc-by-nc-sa","license_terms":"Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction"},{"type":"original","description":"3.8 Implicit Differentiation","author":"Ryan Melton","organization":"","url":"","project":"","license":"cc-by","license_terms":""}],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/261"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/6"}],"version-history":[{"count":10,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/261\/revisions"}],"predecessor-version":[{"id":4530,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/261\/revisions\/4530"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/560"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/261\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=261"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=261"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=261"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=261"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}