{"id":2655,"date":"2021-04-05T18:35:45","date_gmt":"2021-04-05T18:35:45","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=2655"},"modified":"2022-03-16T05:41:02","modified_gmt":"2022-03-16T05:41:02","slug":"logarithmic-differentiation","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/logarithmic-differentiation\/","title":{"raw":"Logarithmic Differentiation","rendered":"Logarithmic Differentiation"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li><span class=\"os-abstract-content\">Use logarithmic differentiation to determine the derivative of a function<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1169737933458\">At this point, we can take derivatives of functions of the form [latex]y=(g(x))^n[\/latex] for certain values of [latex]n[\/latex], as well as functions of the form [latex]y=b^{g(x)}[\/latex], where [latex]b&gt;0[\/latex] and [latex]b\\ne 1[\/latex]. Unfortunately, we still do not know the derivatives of functions such as [latex]y=x^x[\/latex] or [latex]y=x^{\\pi}[\/latex]. These functions require a technique called<strong> logarithmic differentiation<\/strong>, which allows us to differentiate any function of the form [latex]h(x)=g(x)^{f(x)}[\/latex]. It can also be used to convert a very complex differentiation problem into a simpler one, such as finding the derivative of [latex]y=\\frac{x\\sqrt{2x+1}}{e^x \\sin^3 x}[\/latex]. We outline this technique in the following problem-solving strategy.<\/p>\r\n\r\n<div id=\"fs-id1169738197957\" class=\"textbox examples\">\r\n<h3>Problem-Solving Strategy: Using Logarithmic Differentiation<\/h3>\r\n<ol id=\"fs-id1169738197963\">\r\n \t<li>To differentiate [latex]y=h(x)[\/latex] using logarithmic differentiation, take the natural logarithm of both sides of the equation to obtain [latex]\\ln y=\\ln (h(x))[\/latex].<\/li>\r\n \t<li>Use properties of logarithms to expand [latex]\\ln (h(x))[\/latex] as much as possible.<\/li>\r\n \t<li>Differentiate both sides of the equation. On the left we will have [latex]\\frac{1}{y}\\frac{dy}{dx}[\/latex].<\/li>\r\n \t<li>Multiply both sides of the equation by [latex]y[\/latex] to solve for [latex]\\frac{dy}{dx}[\/latex].<\/li>\r\n \t<li>Replace [latex]y[\/latex] by [latex]h(x)[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\nIt may be useful to review your properties of logarithms. These will help us in step 2 to expand our logarithmic function.\r\n<div class=\"textbox examples\">\r\n<h3>Recall: Properties of logarithms<\/h3>\r\n<table summary=\"...\">\r\n<tbody>\r\n<tr>\r\n<td>The Product Rule for Logarithms<\/td>\r\n<td>[latex]{\\mathrm{log}}_{b}\\left(MN\\right)={\\mathrm{log}}_{b}\\left(M\\right)+{\\mathrm{log}}_{b}\\left(N\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>The Quotient Rule for Logarithms<\/td>\r\n<td>[latex]{\\mathrm{log}}_{b}\\left(\\frac{M}{N}\\right)={\\mathrm{log}}_{b}M-{\\mathrm{log}}_{b}N[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>The Power Rule for Logarithms<\/td>\r\n<td>[latex]{\\mathrm{log}}_{b}\\left({M}^{n}\\right)=n{\\mathrm{log}}_{b}M[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>The Change-of-Base Formula<\/td>\r\n<td>[latex]{\\mathrm{log}}_{b}M\\text{=}\\frac{{\\mathrm{log}}_{n}M}{{\\mathrm{log}}_{n}b}\\text{ }n&gt;0,n\\ne 1,b\\ne 1[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<h2><\/h2>\r\n<\/div>\r\n<div id=\"fs-id1169738238112\" class=\"textbook exercises\">\r\n<h3>Example: Using Logarithmic Differentiation<\/h3>\r\n<p id=\"fs-id1169738211866\">Find the derivative of [latex]y=(2x^4+1)^{\\tan x}[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1169738211908\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169738211908\"]\r\n<p id=\"fs-id1169738211908\">Use logarithmic differentiation to find this derivative.<\/p>\r\n\r\n<div id=\"fs-id1169738211912\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lllll} \\ln y &amp; = \\ln (2x^4+1)^{\\tan x} &amp; &amp; &amp; \\text{Step 1. Take the natural logarithm of both sides.} \\\\ \\ln y &amp; = \\tan x \\ln (2x^4+1) &amp; &amp; &amp; \\text{Step 2. Expand using properties of logarithms.} \\\\ \\frac{1}{y}\\frac{dy}{dx} &amp; = \\sec^2 x \\ln (2x^4+1)+\\frac{8x^3}{2x^4+1} \\cdot \\tan x &amp; &amp; &amp; \\begin{array}{l}\\text{Step 3. Differentiate both sides. Use the} \\\\ \\text{product rule on the right.} \\end{array} \\\\ \\frac{dy}{dx} &amp; =y \\cdot (\\sec^2 x \\ln (2x^4+1)+\\frac{8x^3}{2x^4+1} \\cdot \\tan x) &amp; &amp; &amp; \\text{Step 4. Multiply by} \\, y \\, \\text{on both sides.} \\\\ \\frac{dy}{dx} &amp; = (2x^4+1)^{\\tan x}(\\sec^2 x \\ln (2x^4+1)+\\frac{8x^3}{2x^4+1} \\cdot \\tan x) &amp; &amp; &amp; \\text{Step 5. Substitute} \\, y=(2x^4+1)^{\\tan x}.\\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<div class=\"equation unnumbered\">[\/hidden-answer]<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Using Logarithmic Differentiation.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/_nzxTKiFPpo?controls=0&amp;start=816&amp;end=982&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266834\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266834\"]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.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/3.9DerivativesOfExponentialAndLogarithmic816to982_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"3.9 Derivatives of Exponential and Logarithmic Functions\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div id=\"fs-id1169738068346\" class=\"textbook exercises\">\r\n<h3>Example: Using Logarithmic Differentiation<\/h3>\r\n<p id=\"fs-id1169738068356\">Find the derivative of [latex]y=\\large \\frac{x\\sqrt{2x+1}}{e^x \\sin^3 x}[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1169738201884\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169738201884\"]This problem really makes use of the properties of logarithms and the differentiation rules given in this chapter.\r\n<div id=\"fs-id1169738201891\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lllll} \\ln y &amp; = \\ln \\large \\frac{x\\sqrt{2x+1}}{e^x \\sin^3 x} &amp; &amp; &amp; \\text{Step 1. Take the natural logarithm of both sides.} \\\\ \\ln y &amp; = \\ln x+\\frac{1}{2} \\ln (2x+1)-x \\ln e-3 \\ln \\sin x &amp; &amp; &amp; \\text{Step 2. Expand using properties of logarithms.} \\\\ \\frac{1}{y}\\frac{dy}{dx} &amp; = \\frac{1}{x}+\\frac{1}{2x+1}-1-3\\big(\\frac{\\cos x}{\\sin x}\\big) &amp; &amp; &amp; \\text{Step 3. Differentiate both sides.} \\\\ \\frac{dy}{dx} &amp; = y (\\frac{1}{x}+\\frac{1}{2x+1}-1-3 \\cot x) &amp; &amp; &amp; \\text{Step 4. Multiply by} \\, y \\, \\text{on both sides and simplify.} \\\\ \\frac{dy}{dx} &amp; = \\large \\frac{x\\sqrt{2x+1}}{e^x \\sin^3 x} \\normalsize (\\frac{1}{x}+\\frac{1}{2x+1}-1-3 \\cot x) &amp; &amp; &amp; \\text{Step 5. Substitute} \\, y=\\large \\frac{x\\sqrt{2x+1}}{e^x \\sin^3 x}. \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<div class=\"equation unnumbered\">[\/hidden-answer]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738228453\" class=\"textbook exercises\">\r\n<h3>Example: Extending the Power Rule<\/h3>\r\n<p id=\"fs-id1169738228463\">Find the derivative of [latex]y=x^r[\/latex] where [latex]r[\/latex] is an arbitrary real number.<\/p>\r\n[reveal-answer q=\"fs-id1169738228486\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169738228486\"]\r\n<p id=\"fs-id1169738228486\">The process is the same as in the last example, though with fewer complications.<\/p>\r\n\r\n<div id=\"fs-id1169738228494\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lllll} \\ln y &amp; = \\ln x^r &amp; &amp; &amp; \\text{Step 1. Take the natural logarithm of both sides.} \\\\ \\ln y &amp; = r \\ln x &amp; &amp; &amp; \\text{Step 2. Expand using properties of logarithms.} \\\\ \\frac{1}{y}\\frac{dy}{dx} &amp; = r \\frac{1}{x} &amp; &amp; &amp; \\text{Step 3. Differentiate both sides.} \\\\ \\frac{dy}{dx} &amp; = y \\frac{r}{x} &amp; &amp; &amp; \\text{Step 4. Multiply by} \\, y \\, \\text{on both sides.} \\\\ \\frac{dy}{dx} &amp; = x^r \\frac{r}{x} &amp; &amp; &amp; \\text{Step 5. Substitute} \\, y=x^r. \\\\ \\frac{dy}{dx} &amp; = rx^{r-1} &amp; &amp; &amp; \\text{Simplify.} \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<div class=\"equation unnumbered\">[\/hidden-answer]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169737933509\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1169737933517\">Use logarithmic differentiation to find the derivative of [latex]y=x^x[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1169737933537\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169737933537\"]\r\n<p id=\"fs-id1169737933537\">[latex]\\frac{dy}{dx}=x^x(1+\\ln x)[\/latex]<\/p>\r\n\r\n<h4>Hint<\/h4>\r\n<p id=\"fs-id1169737933587\">Follow the problem solving strategy.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWatch the following video to see the worked solution to the above Try It.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/_nzxTKiFPpo?controls=0&amp;start=1160&amp;end=1223&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266833\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266833\"]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.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/3.9DerivativesOfExponentialAndLogarithmic1160to1223_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"3.9 Derivatives of Exponential and Logarithmic Functions\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div id=\"fs-id1169737933594\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1169737933602\">Find the derivative of [latex]y=(\\tan x)^{\\pi}[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1169738233543\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169738233543\"]\r\n<p id=\"fs-id1169738233543\">[latex]y^{\\prime}=\\pi (\\tan x)^{\\pi -1} \\sec^2 x[\/latex]<\/p>\r\n\r\n<h4>Hint<\/h4>\r\n<p id=\"fs-id1169738233597\">Use the result from the last example.<\/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]206261[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li><span class=\"os-abstract-content\">Use logarithmic differentiation to determine the derivative of a function<\/span><\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1169737933458\">At this point, we can take derivatives of functions of the form [latex]y=(g(x))^n[\/latex] for certain values of [latex]n[\/latex], as well as functions of the form [latex]y=b^{g(x)}[\/latex], where [latex]b>0[\/latex] and [latex]b\\ne 1[\/latex]. Unfortunately, we still do not know the derivatives of functions such as [latex]y=x^x[\/latex] or [latex]y=x^{\\pi}[\/latex]. These functions require a technique called<strong> logarithmic differentiation<\/strong>, which allows us to differentiate any function of the form [latex]h(x)=g(x)^{f(x)}[\/latex]. It can also be used to convert a very complex differentiation problem into a simpler one, such as finding the derivative of [latex]y=\\frac{x\\sqrt{2x+1}}{e^x \\sin^3 x}[\/latex]. We outline this technique in the following problem-solving strategy.<\/p>\n<div id=\"fs-id1169738197957\" class=\"textbox examples\">\n<h3>Problem-Solving Strategy: Using Logarithmic Differentiation<\/h3>\n<ol id=\"fs-id1169738197963\">\n<li>To differentiate [latex]y=h(x)[\/latex] using logarithmic differentiation, take the natural logarithm of both sides of the equation to obtain [latex]\\ln y=\\ln (h(x))[\/latex].<\/li>\n<li>Use properties of logarithms to expand [latex]\\ln (h(x))[\/latex] as much as possible.<\/li>\n<li>Differentiate both sides of the equation. On the left we will have [latex]\\frac{1}{y}\\frac{dy}{dx}[\/latex].<\/li>\n<li>Multiply both sides of the equation by [latex]y[\/latex] to solve for [latex]\\frac{dy}{dx}[\/latex].<\/li>\n<li>Replace [latex]y[\/latex] by [latex]h(x)[\/latex].<\/li>\n<\/ol>\n<\/div>\n<p>It may be useful to review your properties of logarithms. These will help us in step 2 to expand our logarithmic function.<\/p>\n<div class=\"textbox examples\">\n<h3>Recall: Properties of logarithms<\/h3>\n<table summary=\"...\">\n<tbody>\n<tr>\n<td>The Product Rule for Logarithms<\/td>\n<td>[latex]{\\mathrm{log}}_{b}\\left(MN\\right)={\\mathrm{log}}_{b}\\left(M\\right)+{\\mathrm{log}}_{b}\\left(N\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>The Quotient Rule for Logarithms<\/td>\n<td>[latex]{\\mathrm{log}}_{b}\\left(\\frac{M}{N}\\right)={\\mathrm{log}}_{b}M-{\\mathrm{log}}_{b}N[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>The Power Rule for Logarithms<\/td>\n<td>[latex]{\\mathrm{log}}_{b}\\left({M}^{n}\\right)=n{\\mathrm{log}}_{b}M[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>The Change-of-Base Formula<\/td>\n<td>[latex]{\\mathrm{log}}_{b}M\\text{=}\\frac{{\\mathrm{log}}_{n}M}{{\\mathrm{log}}_{n}b}\\text{ }n>0,n\\ne 1,b\\ne 1[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h2><\/h2>\n<\/div>\n<div id=\"fs-id1169738238112\" class=\"textbook exercises\">\n<h3>Example: Using Logarithmic Differentiation<\/h3>\n<p id=\"fs-id1169738211866\">Find the derivative of [latex]y=(2x^4+1)^{\\tan x}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169738211908\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169738211908\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738211908\">Use logarithmic differentiation to find this derivative.<\/p>\n<div id=\"fs-id1169738211912\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lllll} \\ln y & = \\ln (2x^4+1)^{\\tan x} & & & \\text{Step 1. Take the natural logarithm of both sides.} \\\\ \\ln y & = \\tan x \\ln (2x^4+1) & & & \\text{Step 2. Expand using properties of logarithms.} \\\\ \\frac{1}{y}\\frac{dy}{dx} & = \\sec^2 x \\ln (2x^4+1)+\\frac{8x^3}{2x^4+1} \\cdot \\tan x & & & \\begin{array}{l}\\text{Step 3. Differentiate both sides. Use the} \\\\ \\text{product rule on the right.} \\end{array} \\\\ \\frac{dy}{dx} & =y \\cdot (\\sec^2 x \\ln (2x^4+1)+\\frac{8x^3}{2x^4+1} \\cdot \\tan x) & & & \\text{Step 4. Multiply by} \\, y \\, \\text{on both sides.} \\\\ \\frac{dy}{dx} & = (2x^4+1)^{\\tan x}(\\sec^2 x \\ln (2x^4+1)+\\frac{8x^3}{2x^4+1} \\cdot \\tan x) & & & \\text{Step 5. Substitute} \\, y=(2x^4+1)^{\\tan x}.\\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div class=\"equation unnumbered\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Using Logarithmic Differentiation.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/_nzxTKiFPpo?controls=0&amp;start=816&amp;end=982&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266834\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266834\" class=\"hidden-answer\" style=\"display: none\">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\/3.9DerivativesOfExponentialAndLogarithmic816to982_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;3.9 Derivatives of Exponential and Logarithmic Functions&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div id=\"fs-id1169738068346\" class=\"textbook exercises\">\n<h3>Example: Using Logarithmic Differentiation<\/h3>\n<p id=\"fs-id1169738068356\">Find the derivative of [latex]y=\\large \\frac{x\\sqrt{2x+1}}{e^x \\sin^3 x}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169738201884\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169738201884\" class=\"hidden-answer\" style=\"display: none\">This problem really makes use of the properties of logarithms and the differentiation rules given in this chapter.<\/p>\n<div id=\"fs-id1169738201891\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lllll} \\ln y & = \\ln \\large \\frac{x\\sqrt{2x+1}}{e^x \\sin^3 x} & & & \\text{Step 1. Take the natural logarithm of both sides.} \\\\ \\ln y & = \\ln x+\\frac{1}{2} \\ln (2x+1)-x \\ln e-3 \\ln \\sin x & & & \\text{Step 2. Expand using properties of logarithms.} \\\\ \\frac{1}{y}\\frac{dy}{dx} & = \\frac{1}{x}+\\frac{1}{2x+1}-1-3\\big(\\frac{\\cos x}{\\sin x}\\big) & & & \\text{Step 3. Differentiate both sides.} \\\\ \\frac{dy}{dx} & = y (\\frac{1}{x}+\\frac{1}{2x+1}-1-3 \\cot x) & & & \\text{Step 4. Multiply by} \\, y \\, \\text{on both sides and simplify.} \\\\ \\frac{dy}{dx} & = \\large \\frac{x\\sqrt{2x+1}}{e^x \\sin^3 x} \\normalsize (\\frac{1}{x}+\\frac{1}{2x+1}-1-3 \\cot x) & & & \\text{Step 5. Substitute} \\, y=\\large \\frac{x\\sqrt{2x+1}}{e^x \\sin^3 x}. \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div class=\"equation unnumbered\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738228453\" class=\"textbook exercises\">\n<h3>Example: Extending the Power Rule<\/h3>\n<p id=\"fs-id1169738228463\">Find the derivative of [latex]y=x^r[\/latex] where [latex]r[\/latex] is an arbitrary real number.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169738228486\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169738228486\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738228486\">The process is the same as in the last example, though with fewer complications.<\/p>\n<div id=\"fs-id1169738228494\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lllll} \\ln y & = \\ln x^r & & & \\text{Step 1. Take the natural logarithm of both sides.} \\\\ \\ln y & = r \\ln x & & & \\text{Step 2. Expand using properties of logarithms.} \\\\ \\frac{1}{y}\\frac{dy}{dx} & = r \\frac{1}{x} & & & \\text{Step 3. Differentiate both sides.} \\\\ \\frac{dy}{dx} & = y \\frac{r}{x} & & & \\text{Step 4. Multiply by} \\, y \\, \\text{on both sides.} \\\\ \\frac{dy}{dx} & = x^r \\frac{r}{x} & & & \\text{Step 5. Substitute} \\, y=x^r. \\\\ \\frac{dy}{dx} & = rx^{r-1} & & & \\text{Simplify.} \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div class=\"equation unnumbered\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169737933509\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1169737933517\">Use logarithmic differentiation to find the derivative of [latex]y=x^x[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169737933537\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169737933537\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169737933537\">[latex]\\frac{dy}{dx}=x^x(1+\\ln x)[\/latex]<\/p>\n<h4>Hint<\/h4>\n<p id=\"fs-id1169737933587\">Follow the problem solving strategy.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/_nzxTKiFPpo?controls=0&amp;start=1160&amp;end=1223&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266833\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266833\" class=\"hidden-answer\" style=\"display: none\">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\/3.9DerivativesOfExponentialAndLogarithmic1160to1223_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;3.9 Derivatives of Exponential and Logarithmic Functions&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div id=\"fs-id1169737933594\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1169737933602\">Find the derivative of [latex]y=(\\tan x)^{\\pi}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169738233543\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169738233543\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738233543\">[latex]y^{\\prime}=\\pi (\\tan x)^{\\pi -1} \\sec^2 x[\/latex]<\/p>\n<h4>Hint<\/h4>\n<p id=\"fs-id1169738233597\">Use the result from the last example.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm206261\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=206261&theme=oea&iframe_resize_id=ohm206261&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-2655\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>3.9 Derivatives of Exponential and Logarithmic Functions. <strong>Authored by<\/strong>: Ryan Melton. <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 class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 1. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\">https:\/\/openstax.org\/details\/books\/calculus-volume-1<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction<\/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":17533,"menu_order":39,"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\":\"3.9 Derivatives of Exponential and Logarithmic Functions\",\"author\":\"Ryan 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