{"id":12445,"date":"2018-08-24t13:24:37","date_gmt":"2018-08-24t20:24:37","guid":{"rendered":"\/\/www.catharsisit.com\/hs\/?p=12445"},"modified":"2018-08-24t13:24:37","modified_gmt":"2018-08-24t20:24:37","slug":"logarithmic-differentiation-ap-calc-review","status":"publish","type":"post","link":"\/\/www.catharsisit.com\/hs\/ap\/logarithmic-differentiation-ap-calc-review\/","title":{"rendered":"what is logarithmic differentiation? ap calc review"},"content":{"rendered":"
what is logarithmic differentiation? basically, it’s a calculus tool that helps you to find derivatives of complicated functions involving a lot of multiplication, division, or powers. this review article walks you through the methods and provides plenty of example problems to help you succeed on the ap calculus exams!<\/p>\n first of all, let’s review what a logarithm<\/strong> is. more specifically, we need to understand how the logarithm function can be used to break down complicated expressions.<\/p>\n you might want to check out the following article before getting started: ap calculus review: properties of exponents and logartithms<\/a>. however, the most important properties for us will be the product<\/strong>, quotient<\/strong>, and power properties<\/strong> for logarithms. here, we focus on a particular logarithm: the natural<\/em> logarithm, ln x<\/em>, though the properties remain true in any base.<\/p>\n in other words, logarithms change…<\/p>\n i like to think of the logarithm as a powerful acid that can dissolve a complicated algebraic expression.<\/p>\n let me illustrate the point with an example.<\/p>\n notice how the original expression involves a huge fraction with roots and powers all over the place. after applying the properties of logarithms, the resulting expression mostly has only plus and minus. (of course, there is a trade-off — there are now three natural logs in the simplified expression.)<\/p>\n now let’s get down to business! how can we exploit these logarithmic simplification rules to help find derivatives?<\/p>\n the most straightforward case is when the function already has a logarithm involved.<\/p>\n find the derivative of first simplify using the properties of logarithms (see work above). then you can take the derivative of each term. but be careful — the final term requires a product rule<\/a>!<\/p>\n in the above example, there was already a logarithm in the function. but what if we want to use logarithmic differentiation when our function has no logarithm?<\/p>\n suppose f<\/em>(x<\/em>) is a function with a lot of products, quotients, and\/or powers. then you might use the method of logarithmic differentiation<\/strong> to find f<\/em>\u00a0‘(x<\/em>).<\/p>\n use logarithmic differentiation to find the derivative of:\u00a0\u00a0 let’s follow the steps outlined above. the first two steps are routine.<\/p>\n on the other hand, step 3 requires us to break down the logarithmic expression using the properties. the work in this step depends on the function. in our case, there is a product of two factors, so we’ll start with the product property. the power property helps to break down the radical. finally, don’t forget the cancellation rule: ln(ex<\/sup><\/em>) = x<\/em>.<\/p>\n next, in steps 4 and 5 apply the derivative and work out the right-hand side.<\/p>\n finally, in step 6, we solve for the unknown derivative by multiplying both sides by y<\/em>. don’t forget to substitute back the original function f<\/em>(x<\/em>) in place of y<\/em>.<\/p>\n a very famous question in calculus class is: what is the derivative of xx<\/sup><\/em> ?<\/p>\n so what is<\/em> the derivative of xx<\/sup><\/em> ?<\/p>\n well, it turns out that only logarithmic differentiation can decide this one for us!<\/p>\n in fact any time there is a function raised to a function power (that is, neither the exponent nor the base is constant), then you will have to use logarithms to break it down before you can take a derivative.<\/p>\n let’s see how it works in the simplest case: xx<\/sup><\/em>.<\/p>\n first, write y<\/em> = xx<\/sup><\/em>.<\/p>\n then, apply the logarithm to both sides:\u00a0\u00a0 ln y<\/em> = ln xx<\/sup><\/em>.<\/p>\n break down the right-hand side of the equation using the algebraic properties of logarithms. in this case, only the power property plays a role.<\/p>\n ln y<\/em> = x<\/em> ln x<\/em><\/p>\n now you can take derivatives of the functions on both sides. but be careful… the function on the right requires a product rule.<\/p>\n (1\/y<\/em>)(dy<\/em>\/dx<\/em>) = (1) ln x<\/em> + x<\/em>(1\/x<\/em>) = ln x<\/em> + 1<\/p>\n finally, multiply both sides by the original function (y<\/em> = xx<\/sup><\/em>) to isolate dy<\/em>\/dx<\/em>.<\/p>\n dy<\/em>\/dx<\/em> = xx<\/sup><\/em>(ln x<\/em> + 1)<\/p>\n and there you have it! the derivative of xx<\/sup><\/em> turns out to be trickier than you might have thought at first, but it’s not impossible.<\/p>\n what is logarithmic differentiation? basically, it’s a calculus tool that helps you to find derivatives of complicated functions involving a lot of multiplication, division, or powers. this review article walks you through the methods and provides plenty of example problems to help you succeed on the ap calculus exams! properties of logarithms first of all, […]<\/p>\n","protected":false},"author":223,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[240],"tags":[241],"ppma_author":[24932],"class_list":["post-12445","post","type-post","status-publish","format-standard","hentry","category-ap","tag-ap-calculus"],"acf":[],"yoast_head":"\nproperties of logarithms<\/h2>\n
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using the properties<\/h3>\n
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logarithmic differentiation<\/h2>\n
example 1<\/h3>\n
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solution<\/h4>\n
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the general method<\/h3>\n
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example 2<\/h3>\n
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solution<\/h4>\n
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functions raised to a function power<\/h2>\n
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summary<\/h2>\n
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